I borrowed the ideas from the following books.
Abelian categories with application to rings and modules by Popescu, 1973.
Theory of categories by Mitchell, 1964.
Notations and Conventions We fix a Grothendieck universe $\mathcal{U}$. We consider only categories which belong to $\mathcal{U}$. Let $\mathcal{C}$ be a category. We denote by Ob($\mathcal{C}$) the set of objects of $\mathcal{C}$. Often, by abuse of notation, we use $\mathcal{C}$ instead of Ob($\mathcal{C}$). We denote by Mor($\mathcal{C}$) the set of morphisms of $\mathcal{C}$.
Let $f:X \rightarrow Y$ be a morphism of $\mathcal{C}$. We denote by dom($f$) the domain of $f$, i.e. $X$ = dom($f$). We denote by codom($f$) the codomain of $f$, i.e. $Y$ = codom($f$).
Definition Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let $I$ be a small set. Let $(X_i)_I$ be a family of subobjects of $X$. If $(X_i)_I$ satisfies the following condition, $(X_i)_I$ is called a directed family of subobjects of $X$.
For any $i, j \in I$, there exists $k \in I$ such that $X_i \subset X_k$ and $X_j \subset X_k$.
Lemma 1 Let $\mathcal{A}$ be a cocomplete abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $A$ = colim $F$. For each $i \in I$, let $f_i:F(i) \rightarrow A$ be the canonical morphism. For each $i \in I$, let $A_i$ = Im($f_i$). Since $\mathcal{A}$ is cocomplete, $\sum A_i$ exists. Then $A = \sum A_i$.
Proof: Let $B = \sum A_i$. Let $m:B \rightarrow A$ be the canonical monomorphism. Since $A_i$ = Im($f_i$) for each $i \in I$, there exists $g_i:F(i) \rightarrow B$ such that $f_i = mg_i$. Let $u: i \rightarrow j$ be a morphism of I. Since $f_i = f_jF(u)$, $mg_i = mg_jF(u)$. Since $m$ is a monomorphism, $g_i = g_jF(u)$. Hence there exists $g:A \rightarrow B$ such that $g_i = gf_i$ for each $i$. Hence $mgf_i = mg_i = f_i$ for each $i$. Hence $mg = 1_A$. Hence $A \subset B$. Hence $A = B$. QED
Lemma 2 Let $\mathcal{C}$ be a cocomplete category. Let I be a small category. Let $\mathcal{C}^I$ be the category of functors: $I \rightarrow \mathcal{C}$. Then colim$: \mathcal{C}^I \rightarrow \mathcal{C}$ preserves colimits.
Proof: Let $\Delta: \mathcal{C} → \mathcal{C}^I$ be the diagonal functor, i.e. for each $X \in \mathcal{C}$ and for each $i \in I$, $\Delta(X)(i) = X$. Since colim is a left adjoint functor of $\Delta$, it preserves colimits(MacLane: Categories for the working mathematician, Chapter V, Section 5, Theorem 1, p.114). QED
Lemma 3 Let $\mathcal{A}$ be a cocomplete abelian category which satisfies (AB5). Let $X$ be an object of $\mathcal{A}$. Let I be a small filtered category. Let Sub($X$) be the category of subobjects of $X$. Let $F: I \rightarrow$ Sub($X$) be a functor.
Then $\sum F(i)$ = colim $F$.
Proof: For each $i \in I$, Let $u_i:F(i) \rightarrow$ colim $F$ be the canonical morphism. For each $i \in I$, Let $m_i:F(i) \rightarrow X$ be the canonical monomorphism. Since $(m_i)_I$ is a cocone, it induces a morphism $f$: colim $F \rightarrow X$. By (AB5), $f$ is mono. Hence we can regard colim $F$ as a subobject of $X$. Since $fu_i = m_i$ for each $i$, $F(i) \subset$ colim $F$.
Let $Z$ be a subobject of $X$. Let $r: Z \rightarrow X$ be the canonical monomorphism. Suppose $F(i) \subset Z$ for each $i$. Let $k_i: F(i) \rightarrow Z$ be the canonical monomorphism. Since $(k_i)_I$ is a cocone, it induces a morphism $g$: colim $F \rightarrow Z$. For each $i \in I$, $rgu_i = rk_i = m_i$. Hence $f = rg$. Hence colim $F \subset Z$. QED
Lemma 4 Let $\mathcal{A}$ be a cocomplete abelian category. Let $f:X \rightarrow Y$ be a morphism of $\mathcal{A}$. Let $I$ be a small set. Let $(X_i)_I$ be a family of subobjects of $X$. Then $\sum f(X_i) = f(\sum X_i)$.
Proof: For each $i \in I$, $X_i \subset \sum X_i$. Hence $f(X_i) \subset f(\sum X_i)$.
Let $Z$ be a subobject of $X$. Suppose $f(X_i) \subset Z$ for each $i$. Then $f^{-1}(f(X_i)) \subset f^{-1}(Z)$. Since $X_i \subset f^{-1}(f(X_i))$, $X_i \subset f^{-1}(Z)$. Hence $\sum X_i \subset f^{-1}(Z)$. Hence $f(\sum X_i) \subset f(f^{-1}(Z)) \subset Z$. QED
Lemma 4.5 Let $\mathcal{A}$ be a cocomplete abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $X$ = colim $F$. Let $(s_i: F(i) \rightarrow Y)_I$ be a cocone. Let $f:X \rightarrow Y$ be the morphism induced by the cocone. Then $f(X) = \sum s_i(F(i))$.
Proof: For each $i$, let $u_i: F(i) \rightarrow X$ be the canonical morphism. For each $i$, $fu_i = s_i$. Hence $f(u_i(F(i)) = s_i(F(i)) \subset f(X)$.
Let $Z$ be a subobject of $Y$. Suppose $s_i(F(i)) \subset Z$ for each $i$. For each $i$, $s_i$ induces $t_i: F(i) \rightarrow Z$. Since $(t_i: F(i) \rightarrow Z)_I$ is a cocone, it induces $g:X \rightarrow Z$. Let $m: Z \rightarrow Y$ be the canonical monomorphism. $mgu_i = mt_i = s_i$ for each i. Hence $f = mg$. Hence $f(X) \subset Z$. QED
Lemma 5 Let $\mathcal{A}$ be a cocomplete abelian category. Let $X$ be an object of $\mathcal{A}$. Let $I$ be a small set. Let $(X_i)_I$ be a family of subobjects of $X$. Then $\bigoplus X/X_i$ = $X/(\sum X_i)$.
Proof: For each $i \in I$, the following sequence is exact. $0 \rightarrow X_i \rightarrow X \rightarrow X/X_i \rightarrow 0$. By Lemma 2, colim preserves cokernels. Hence, colim $X_i \rightarrow X \rightarrow$ colim $X/X_i \rightarrow 0$ is exact. By Lemma 4.5, Im(colim $X_i \rightarrow X$) = $\sum X_i$. Hence colim $X/X_i$ = $X/(\sum X_i)$. QED
Lemma 5.4 Suppose the following is a pullback diagaram in an abelian category. $\begin{matrix} A&\stackrel{f}{\rightarrow}&B\\ \downarrow&&\downarrow\\ C&\stackrel{h}{\rightarrow}&D \end{matrix} $
Suppose the following sequence is exact. $0 \rightarrow C \stackrel{h}{\rightarrow}D \rightarrow E$
Then $0 \rightarrow A \stackrel{f}{\rightarrow}B \rightarrow E$ is exact.
Proof: Left to the readers.
Lemma 5.5 Consider the following commutative diagram with two horizontal exact sequences in an abelian category.
$X \rightarrow Y \rightarrow Z \rightarrow 0$
$0 \rightarrow X' \rightarrow Y' \rightarrow Z' \rightarrow 0$
Suppose the left square is a pullback. Then $Z \rightarrow Z'$ is mono.
Proof: We call s the above morphism $Z \rightarrow Z'$.
Let $r:T \rightarrow Z$ be a morphism such that sr = 0.
There exists the following pullback diagaram. $\begin{matrix} P&\stackrel{u}{\rightarrow}&T\\ \downarrow&&\downarrow{r}\\ Y&\stackrel{}{\rightarrow}&Z \end{matrix} $
By Lemma 5.4, $0 \rightarrow X \rightarrow Y \rightarrow Z’$ is exact.
Hence there exists $P \rightarrow X$ such that $P \rightarrow Y = P \rightarrow X \rightarrow Y$.
Hence $ru$ = 0. On the other hand, since a pullback of an epimorphism in an abelian category is epi(MacLane Proposition 2, p.199), $u$ is epi. Hence $r$ = 0. QED
Lemma 6 Let $\mathcal{A}$ be an abelian category. Let $f:X \rightarrow Y$ be a morphism of $\mathcal{A}$. Let $Z \subset Y$. Then $X/f^{-1}(Z)$ is canonically isomorphic to $f(X)/(f(X) \cap Z)$.
Proof:
Consider the following commutative diagram with two horizontal exact sequences.

By Lemma 5.5, $X/f^{-1}(Z) \rightarrow f(X)/(f(X) \cap Z)$ is mono. Since $X \rightarrow f(X)$ is epi, $X \rightarrow X/f^{-1}(Z) \rightarrow f(X)/(f(X) ∩ Z)$ is epi. Hence $X/f^{-1}(Z) \rightarrow f(X)/(f(X) \cap Z)$ is epi. Hence $X/f^{-1}(Z) \rightarrow f(X)/(f(X) \cap Z)$ is an isomorphism. QED
Note If you are willing to accept Mitchell's embedding theorem, Lemma 6 will be trivial.
Lemma 7 Let $\mathcal{A}$ be a cocomplete abelian category. Suppose $\mathcal{A}$ has the following property.
Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.
Let $f:Y \rightarrow X$ be a morphism of $\mathcal{A}$. Let $(X_i)_I$ be a directed family of subobjects of $X$. Then, $f^{-1}(\sum X_i) = \sum f^{-1}(X_i)$.
Proof: By Lemma 6, for each $i$, $Y/f^{-1}(X_i)$ is canonically isomorphic to $f(Y)/(f(Y) \cap X_i)$.
Hence $\bigoplus Y/f^{-1}(X_i)$ is canonically isomorphic to $\bigoplus f(Y)/(f(Y) \cap X_i)$.
By Lemma 5, $Y/\sum f^{-1}(X_i)$ = $\bigoplus Y/f^{-1}(X_i)$.
Hence $Y/\sum f^{-1}(X_i)$ = $\bigoplus f(Y)/(f(Y) \cap X_i)$.
By Lemma 5, $\bigoplus f(Y)/(f(Y) \cap X_i)$ = $f(Y)/\sum (f(Y) \cap X_i)$.
By the assumption, $f(Y)/\sum (f(Y) \cap X_i)$ = $f(Y)/((\sum X_i) \cap f(Y))$.
By Lemma 6, $Y/f^{-1}(\sum X_i)$ is canonically isomorphic to $f(Y)/((\sum X_i) \cap f(Y))$.
Hence $Y/\sum f^{-1}(X_i)$ is canonically isomorphic to $Y/f^{-1}(\sum X_i)$.
Hence $f^{-1}(\sum X_i)$ = $\sum f^{-1}(X_i)$. QED
Lemma 7.3 Let $\mathcal{C}$ be a category. Let $X$ be an object of $\mathcal{C}$. Let Sub($X$) be the category of subobjects of $X$. Let $I$ be a small set. Let $(X_i)_I$ be a directed family of subobjects of $X$. Then there exists a preorder on $I$ making $I$ a filtered category and a functor $F: I \rightarrow$ Sub($X$) such that $F(i) = X_i$ for each $i \in I$.
Proof: Define $i \leq j$ if and only if $X_i \subset X_j$. QED
Lemma 7.5 Let $\mathcal{A}$ be an abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $i \in I$. Let $(i\downarrow I)$ be the coslice category under i. Let Sub($F(i)$) be the category of subobjects of $F(i)$. Then there exists a functor $G$: $(i\downarrow I) \rightarrow$ Sub($F(i)$) such that $G(u)$ = Ker($F(u)$) for each $u \in (i\downarrow I)$.
Proof:Clear.
Lemma 8 Let $\mathcal{A}$ be an abelian category. Let $I$ be a small filtered category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $i \in I$. Let $J$ = {$u \in$ Mor($I$); $i$ = dom($u$)}. Then (Ker($F(u))$)$_J$ is a directed family of subobjects of $F(i)$.
Proof: Let $(i\downarrow I)$ be the coslice category under i. $(i\downarrow I)$ is clearly a filtered category. Since $J$ = Ob($(i\downarrow I)$), the assertion follows immediately from Lemma 7.5. QED
Lemma 8.5 Let $\mathcal{A}$ be a cocomplete abelian category. Let $I$ be a small category. Let $F: I \rightarrow \mathcal{A}$ be a functor. Let $S = \bigoplus_i F(i)$, where $i$ runs over every object of $I$. Let $m_i: F(i) \rightarrow S$ be the canonical monomorphism for each $i \in I$. Let $M$ = $\sum_u$ Im($m_i - m_jF(u)$), where $u$ runs over every morphism of $I$ and $i$ = dom($u$), $j$ = codom($u$). Let $\pi:S \rightarrow S/M$ be the canonical epimorphism. Let $f_i = \pi m_i$ for each $i \in I$. Then $S/M$ = colim $F$ with canonical morphisms $f_i: F(i) \rightarrow S/M$ for each $i \in I$.
Proof:Left to the readers.
Lemma 8.6 Let $I$ be a filtered category. Let $V$ be a non-empty finite subset of Ob($I$). Let $T$ be a finite subset of Mor($I$) such that dom($u$) $\in V$ and codom($u$) $\in V$ whenever $u \in T$. Then there exists $p \in$ Ob($I$) and a morphism $f_i: i \rightarrow p$ for each $i \in V$ with the following property.
For each $u:i \rightarrow j$ in $T$, $f_i = f_ju$.
Proof: There exists $q \in I$ such that there exists a morphism $g_i:i \rightarrow q$ for each $i \in V$. Let $u:i \rightarrow j$ in $T$. There exists $r_u \in I$ and a morphism $h_u:q \rightarrow r_u$ such that $h_ug_i = h_ug_ju$. There exists $r \in I$ such that there exists a morphism $r_u \rightarrow r$ for each $u \in T$.
Hence, for each $u:i \rightarrow j$ in $T$ there exist a morphism $g_{u, i}: i \rightarrow r$ and a morphism $h_{u, j}: j \rightarrow r$ such that $g_{u, i} = h_{u, j}u$.
For each $i \in V$, let $G_i$ be the set {$g_{u, i}: i$ = dom($u$), $u \in T$}, and let $H_i$ be the set {$h_{u, i}: i$ = codom($u$), $u \in T$}. Let $S_i = G_i \cup H_i$ for each $i \in V$. By the properties of a filtered category, we can assume that $S_i$ consists of one morphism $f_i$ with a common codomain $p$ for each $i \in V$. If $S_i$ is empty, we can assume that there exists a morphism $f_i:i \rightarrow p$ which has no condition.
QED
Lemma 9 Let $\mathcal{A}$ be a cocomplete abelian category. Suppose $\mathcal{A}$ has the following property.
Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.
Let I be a small filtered category. Let $F: I \rightarrow \mathcal{A}$ be a functor. For each $i$, let $f_i: F(i) \rightarrow$ colim($F$) be the canonical morphism. Then, for each $i$, Ker($f_i$) = $\sum$ Ker($F(u)$), where $u$ runs over every morphism such that $i$ = dom($u$).
Proof: We use the notations of Lemma 8.5. Let $T$ be a subset of Mor($I$). Let $M_T$ = $\sum_{u \in T}$ Im($m_i - m_jF(u)$), where $i$ = dom($u$), $j$ = codom($u$). Then $M = \sum_T M_T$, where T runs through all finite subsets of Mor($I$). Hence, by Lemma 8.5 and Lemma 7, Ker($f_i$) = $m_i^{-1}(M)$ = $\sum_T m_i^{-1}(M_T)$, where T runs through all finite subsets of Mor($I$). It suffices to prove: For each finite subset $T$ of Mor($I$), $m_i^{-1}(M_T) \subset$ Ker($F(u)$) for some $u \in$ Mor($I$) such that $i$ = dom($u$).
Let $V$ be the set of $k \in I$ such that $k$ = $i$ or $k$ = dom($u$) or $k$ = codom($u$) for some $u \in T$.
Since $I$ is filtered, by Lemma 8.6, there exists $p \in$ Ob($I$) and a morphism $v_k: k \rightarrow p$ for each $k \in V$ with the following property.
For each $u \in$ Mor($I$) such that k = dom($u$) $\in V$ and $j$ = codom($u$) $\in V$, $v_k = v_ju$.
We define $f:S \rightarrow F(p)$ as follows. Let $k$ be any object of $I$. If $k \in V$, $fm_k = F(v_k)$, otherwise $fm_k = 0$.
For each $u \in T$, let $k$ = dom($u$), $j$ = codom($u$). Then $f(m_k - m_jF(u))$ = $F(v_k) - F(v_j)F(u)$ = $0$. Hence, by Lemma 4, $f(M_T)$ = $0$. Since $m_i(m_i^{-1}(M_T)) \subset M_T$, $0$ = $f(m_i(m_i^{-1}(M_T)))$ = $F(v_i)(m_i^{-1}(M_T))$. Hence $m_i^{-1}(M_T) \subset$ Ker($F(v_i)$) as required. QED
Proposition 1 Let $\mathcal{A}$ be a cocomplete abelian category. Suppose $\mathcal{A}$ has the following property.
Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.
Then $\mathcal{A}$ satisfies (AB5).
Proof: Let $I$ be a small filtered category. By Lemma 2, colim$: \mathcal{A}^I \rightarrow \mathcal{A}$ preserves colimits. In particular, it preserves cokernels. Hence it is right exact. It suffices to prove that it preserves monomorphisms.
Let $f: F \rightarrow G$ be a monomorphism of $\mathcal{A}^I$. Let $K$ = Ker(colim($f$)). For each i, let $u_i: F(i) \rightarrow$ colim $F$ be the canonical morphism. Let $A_i$ = $u_i(F(i))$ for each i. Since $I$ is a filtered category, $(A_i)_I$ is a directed family of subobjects of colim $F$. By Lemma 1, colim $F$ = $\sum A_i$. By the assumption, $K$ = $(\sum A_i) \cap K = \sum (A_i \cap K)$. Suppose $K \neq 0$. There exists $k \in I$ such that $A_k \cap K \neq 0$. Since $A_k$ = Im($u_k$), $u_k^{-1}(A_k \cap K) \neq 0$. Let $M = u_k^{-1}(A_k \cap K)$. Then $u_k(M) \neq 0$. For each i, let $v_i: G(i) \rightarrow$ colim $G$ be the canonical morphism. $v_k(f_k(M))$ = (colim $f$)($u_k(M)$) = (colim $f$)($A_k \cap K$) $\subset$ (colim f)($K$) = $0$. Hence $f_k(M) \subset$ Ker($v_k$). By Lemma 9, $f_k(M) \subset \sum$ Ker($G(t)$), where $t$ runs over every morphism such that $k$ = dom($t$).
By Lemma 8 and the assumption, $f_k(M)$ = $\sum$ (Ker($G(t)$) $\cap f_k(M)$). Since $f_k$ is mono, $M$ = $f_k^{-1}(f_k(M))$. By Lemma 7, $M$ = $f_k^{-1}(f_k(M))$ = $f_k^{-1}(\sum$ (Ker($G(t)$) $\cap f_k(M)))$ = $\sum f_k^{-1}$(Ker($G(t)$ $\cap f_k(M))$. For each morphism $t: k \rightarrow j$, Let $N_t = f_k^{-1}$(Ker($G(t)$) $\cap f_k(M))$. Then $G(t)f_k(N_t) = 0$. Since $G(t)f_k = f_jF(t)$, $f_jF(t)(N_t) = G(t)f_k(N_t) = 0$. Since $f_j$ is mono, $F(t)(N_t) = 0$. Hence, by Lemma 9, $u_k(N_t) = 0$. Hence, by Lemma 4, $u_k(M) = u_k(\sum N_t) = \sum u_k(N_t) = 0$. This is a contradiction. QED
Proposition 2 Let $\mathcal{A}$ be a cocomplete abelian category satisfying (AB5). Let $A$ be an object of $\mathcal{A}$. Let $(A_i)_I$ be a directed family of subobjects of $A$. Then, for every subobject $B$ of $A$, $(\sum A_i) \cap B = \sum (A_i \cap B)$.
Proof: Let $C = \sum A_i$. For each i, we have the following exact sequence.
$0 \rightarrow A_i \cap B \rightarrow A_i \rightarrow C/(C \cap B)$
By (AB5), Lemma 7.3 and Lemma 3, we get the following exact sequence.
$0 \rightarrow \sum (A_i \cap B) \rightarrow C \rightarrow C/(C \cap B)$
Hence $(\sum A_i) \cap B = \sum (A_i \cap B)$. QED