You don't need projectivity of $E$ because you have sufficiently many push-out diagrams and short exact sequences around.
Let us fix the notation first. Form the push-out under $i$ and $f$ to obtain the diagram
i i' B >---> E --->> A | | | |f |f' |1 V j V j' v C >---> P --->> A
and form the push-out under $i$ and $g$ to obtain the diagram
i i' B >---> E --->> A | | | |g |g' |1 V k V k' v C >---> Q --->> A.
We want to show that the extensions C >-> P ->> A
and C >-> Q ->> A
are equivalent if and only if there exists a morphism m: E -> C
such that mi = g - f
.
To say that C >-> P ->> A
and C >-> Q ->> A
are equivalent is to say that there is an isomorphism l: P -> Q
such that
j j' C >---> P --->> A | | | |1 |l |1 V k V k' v C >---> Q --->> A
commutes. Consider the morphism g'-lf': E -> Q
. We have
k'(g'-lf') = k'g' - k'lf' = i' - j'f' = 0
so that (g'-lf') = km
for a unique morphism m: E -> C
since k = ker k'
. Now notice that
kmi = (g' - lf')i = g'i - lf'i = kg - ljf = k (g - f)
so that the fact that k
is a monomorphism gives that mi = g-f
, as we wanted.
Conversely, suppose there is a morphism m: E -> C
such that mi = g-f
. Then
kf = k (g - mi) = kg - kmi = g'i - kmi = (g' - km) i
so that we have a commutative square
i B >---> E | | |f |g'-km V k V C >---> Q
Applying the push-out property of BECP
to that square we get a unique morphism l: P -> Q
such that k = lj
and g'-km = lf'
and I claim that the diagram
j j' C >---> P --->> A | | | |1 |l |1 V k V k' v C >---> Q --->> A
commutes. For the left hand square this is given and for the right hand square notice that
(k'l - j') f' = k'lf' - j'f' = k' (g' - km) - i' = k'g' - i' = 0
and
(k'l - j') j = k'lj = k'k = 0
so that (k'l - j')[j f'] = [0 0]
and this implies that k'l-j' = 0
because the morphism [j f']: C+E -->> P
is an epimorphism, as BECP
is a push-out by hypothesis. It follows from the five lemma that the two extensions C >-> P ->> A
and C >-> Q ->> B
are equivalent.
It is a good exercise to show that a similar argument applied to an appropriate square (I'll let you find the morphism (?)
yourself)
i B >---> E | | |g |(?) V j V C >---> P
and exploiting the push-out property of BECQ
yields a morphism l': Q -> P
which is inverse to l: P -> Q
.