 Solution

Let $S$ be a finite generating set of $H$, choose one lift $\tilde{s}\in G$ for every $s\in S$, and put $K=\ker f$. The finite setgenerates $G$: lift a word representing $f(g)$ and observe that the discrepancy from $g$ lies in $K$.

Use this generating set for $G$. The quotient map does not increase word length, while lifting a shortest word in $H$ leaves only a final element of $K$, whose word length is at most one. HenceThe map is surjective, so it is a finite-kernel quotient quasi-isometry. Therefore $G$ is finitely generated and quasi-isometric to $H$.