Edit: We first check that $\bigcup M_i$ is an $A$ - module: We will check the only non-trivial part here which is closure under addition. Suppose that we have $y,z \in \bigcup M_i$. Then there exist $j,k$ such that $y \in M_j$ and $z \in M_k$. Now by assumption there exists $l \in I$ such that $M_j + M_k \subseteq M_l$. Hence it is clear that $y+z \in M_l \subset \bigcup M_i$ showing that $\bigcup M_i$ is closed under addition.
I don't think your proof is quite right because you can't just say "concatenation of homomorphisms": how do you define $i^{-1}(m)$? I think your proof breaks down here because you cannot just do that. The way I would do it is as follows:
We want to prove that whenever we have maps $f_i : M_i \rightarrow N$ where $N$ is some $A$ - module and $f_i = f_j \circ \mu_{ij}$ whenever $i\leq j$, there is a unique $A$ - module homomorphism $L : \bigcup M_i \to N$. Take an element $x \in \bigcup M_i$. Then $x$ is in at least one $M_j$, so we can define $L(x)$ to be equal to $f_j(x)$. We need to check that this is well defined. Suppose $x$ is also in some other $M_k$. Then we know that we can choose an $l \in I$ such that $l \geq j,k$. By definition of $f_l$ we have that $f_j(x) = f_l\circ \mu_{jl}(x)$. But the right hand side is just $x$ viewed as an element of $M_l$ and then mapped under $f_l$. Similarly writing down $f_k(x) = f_l \circ \mu_{kl}(x)$, the right hand side is just $x$ viewed as an element of $M_l$ and then mapped under $f_l$. It follows that $f_j(x) = f_k(x)$ so that $L$ is well defined.
We now check that $L$ is an $A$ - module homomorphism; the only non-trivial part to check is that $L$ is additive. So suppose we have $x,y \in \bigcup M_i$. Then there is $j,k$ such that $x \in M_j$ and $y \in M_k$. Then we know that there exists an $l \geq j,k$ such that $x+y \in M_l$ and so $L(x+y) = f_l (x+y)$. Now $L(x) = f_j(x)$, $L(y) = f_k(y)$. Since
$f_j(x) = f_l\circ \mu_{jl}(x), \hspace{4mm} f_k(y) = f_l\circ \mu_{kl}(y)$
we can say that $f_j(x) + f_k(y) = f_l(x) +f_l(y)$. This is because $\mu_{jl}(x)$ and $\mu_{kl}(y)$ just views $x$ and $y$ respectively as elements of $M_l$. It follows that $L$ is additive. It now remains to check that $L$ is uniquely determined by the $f_i$. Suppose we have another map $\varphi : \bigcup M_i \longrightarrow N$ such that $\varphi \circ \tau_i = f_i$. Then it follows that
$\varphi \circ \tau_i(x) = L \circ \tau_i(x)$
given any $i \in I$ and $x \in M_i$. But then since the $\tau_i$ are inclusion maps this is effectively saying that $L(x) = \varphi(x)$ for all $x \in M_i$ for any $M_i$. Since $x$ is arbitrary we conclude that $L = \varphi$ showing uniqueness. This completes the proof that
$\varinjlim M_i \cong \bigcup M_i.$
Edit: I think you have your universal properties wrong: The universal property of the direct limit is this: If you have $N$ an $A$ - module and for each $i \in I$ let $\alpha_i : M_i \to N$ be an $A$ - module homomorphism such that $\alpha_i = \alpha_j \circ \mu_{ij}$ whenver $i \leq j$. The there exists a unique homomorphism $\alpha : M \to N$ such that $\alpha_i = \alpha \circ \mu_i$ for all $i \in I$.
Edit: Let me show you why it suffices to show that $\bigcup M_i$ satisfies the universal property of $\varinjlim M_i$ to prove that $\bigcup M_i \cong \varinjlim M_i$. Now we have inclusion maps $\tau_i : M_i \to \bigcup M_i$ such that clearly $\tau_i = \tau_j \circ \mu_{ij}$ for any $i \leq j$ because the $\mu_{ij}$ are just inclusion maps. So now we just need to show that $\bigcup M_i$ has the property that given compatible $A$ - module homomorphisms $f_i : M_i \to N$ for some $A$ - module $N$ such that $f_i = f_j \circ \mu_{ij}$ whenever $i \leq j$, there is a unique $A$ - linear map $L : \bigcup M_i \to N$ such that
$f_i = L \circ \tau_i \hspace{3mm} \forall i\in I.$
To see why we only need to verify this, suppose the result above about $\bigcup M_i$ holds. Then putting in place of $f_i$ the usual maps $\phi_i$ from $M_i$ to $\varinjlim M_i$ and $N = \varinjlim M_i$, we have a unique linear map $L : \bigcup M_i \longrightarrow \varinjlim M_i$ such that
$\phi_i = L \circ \tau_i \hspace{3mm} \forall i \in I.$
Recall how the maps $\phi_i$ are defined. We have $M_i \stackrel{\lambda_i}{\longrightarrow} \bigoplus_{i\in I} M_i \stackrel{ \mu}{\longrightarrow} \varinjlim M_i$ so
$\phi_i \stackrel{\text{def}}{\equiv} \mu \circ \lambda_i$
where $\mu$ is the canonical projection from the direct sum onto the direct limit, and the $\lambda_i$ the canonical injections from $M_i$ into the direct sum. Now because we have maps $\tau_i$ out of $M_i$ to $\bigcup M_i$ such that $\tau_i = \tau_j \circ \mu_{ij}$ whenever $i\leq j$, then by the universal property of the direct limit, there is a unique $A$ - module homomorphism $L': \varinjlim M_i \rightarrow \bigcup M_i$ such that
$\tau_i = \phi_i \circ L' \hspace{3mm} \forall i \in I.$
From here it is not hard to see that $L$ and $L' $ are mutual inverses so that $\bigcup M_i \cong \varinjlim M_i$.
$\hspace{6in} \square$