The modified question has a positive answer if $N$ is finitely generated.

Consider an extension $1 \to N \to G \to \mathbb Z \to 1$ and take a lift $u \in G$ of the generator of $\mathbb Z$. If $N$ is finitely generated and $H' \subset N$ is a subgroup of finite index, then the intersection of all subgroups of index $[H':N]$ (call it $H$) is invariant under conjugation by $u$. Hence, for all $m \in \mathbb Z$, the subgroup $Hu^{m \mathbb Z}$ is a normal in $G$finite index normal subgroup of $G$.

Hence, if $N$ is finitely generated and residually finite, then $G$ is residually finite as well.