Let $f: A\rightarrow B$ be a surjective homomorphism between abelian groups. I want to find such subgroup $B'\subset A$ that $\phi=f|_{B'}:B'\rightarrow B$ would be an isomorphism.
I think it's easy to get if $B$ is a free abelian group, you just need to find preimages $\{e'_i\}$ of all generators $\{e_i\}$, generate a subgroup and that would be $B'$, also $g(e_i)=e'_i$ would define the inverse.
But is it possible to get such $B'$ if $B$ is not free, using only surjectivity?
This is impossible (generally). Consider the natural map $\phi: \mathbb{Z} \to \mathbb{Z}_2$. This is surjective, but $\mathbb{Z}$ has no elements of order $2$ (or, equivalently, subgroups of order $2$). I don't see how the non-finite-generation of $B$ could ever help.