Group homomorphism $f$ is surjective iff $g$ is Let $G$ be an additive group, and let $u, v:G\to G$ to be two endomorphisms. Define $f(x) = x- v(u(x))$ and $g(x) = x-u(v(x))$. The question is to show that $f$ is surjective iff $g$ is. 
I'm only able to show that $u:\ker f\cong \ker g$, but unable to show the statement needed. Observe that $u(f(x)) = g(u(x))$ and $v(g(x)) = f(v(x))$.

We have the following general fact: In a ring $R$ (not assumed to be commutative) with elements $u,v$, if $1-uv$ is invertible, then $1-vu$ is invertible. In fact, one checks that 
$(1-vu)^{-1} = 1 + v (1-uv)^{-1} u$
(This formula has a nice motivation using the geometric series.)
We may apply this to the ring $R=\mathrm{End}(G)$ for an abelian group $G$ (with $1=\mathrm{id}_G$) and see that $1-uv$ is an isomorphism if and only if $1- v u$ is an isomorphism.
But we can also use the formula above to prove that $1-vu$ is surjective when $1-uv$ is surjective:
Let $g \in G$. Choose some $h \in G$ with $(1-uv)(h) = u(g)$, i.e. $h - u(v(h))=u(g)$. Then $g + v(h) \in G$ satisfies
$$(1-vu)(g + v(h))=g + v(h) - v(u(g+v(h)))=g + v(h - u(v(h)) - u(g))=g.$$
This proof also works when $G$ is not abelian (where $x-y$ abbreviates $x+(-y)$ and we use $-(x+y)=(-y)+(-x)$ in the calculation).
