From the comments, it seems that the actual situation is:
Suppose that $f,h\colon\mathbb{Z}\oplus\mathbb{Z}\to\mathbb{Z}\oplus\mathbb{Z}$ are group homomorphisms, and that $h$ is a retraction of $f$; that is, $h\circ f = \mathrm{id}$. Is $f$ an isomorphism?
The answer in that situation is "yes".
Since $h\circ f$ is bijective, $f$ is one-to-one and $h$ is onto. It is also straightforward to verify that $\mathbb{Z}\oplus\mathbb{Z}\cong \mathrm{Im}(f) \oplus \mathrm{ker}(h).$ Both $\mathrm{Im}(f)$ and $\mathrm{ker}(h)$ are free abelian. Since $f$ is one-to-one, $\mathrm{Im}(f)$ is free abelian of rank $2$. Since the rank of a free abelian group is uniquely determined, this means that $2 = \mathrm{rank}(\mathbb{Z}\oplus\mathbb{Z}) = \mathrm{rank}(\mathrm{Im}(f))+\mathrm{rank}(\mathrm{ker}(h)) = 2 + \mathrm{ker}(h),$ so $\mathrm{ker}(h)$ is trivial. Therefore, $h$ is one-to-one and onto, hence bijective, and since $h\circ f = \mathrm{id}$, it follows that $f$ is the inverse of $h$, and an isomorphism.