Just slightly changing the notation: the question asks, given a group with normal subgroup $N$ and quotient $Q=G/N$, such that $N$ and $Q$ are abelian and residually finite, whether $G$ has to be residually finite.
(1) One counterexample is $K\rtimes K^*$ when $K$ is an infinite field of characteristic $p>0$. This is not residually finite (because $K$ is mapped injectively or trivially in every quotient), while $K$ is residually finite.
The group $K^*$ may fail to be residually finite (e.g., when $K$ is algebraically closed), but for $K=\mathbf{F}_p(t)$ it is residually finite (direct product of $\mathbf{F}_p^*$ with a free abelian group of infinite rank).
(2) Here's a counterexample, also split extension, with $Q$ finitely generated (infinite cyclic). Consider $M=\mathbf{F}_p[t^{\pm 1}]/\mathbf{F}_p[t]$, with $T$ the surjective locally nilpotent endomorphism induced by multiplication by $t$. Write $U=\mathrm{Id}+T$. It is thus invertible and we can consider the corresponding semidirect product $G=M\rtimes_U\mathbf{Z}$. Since for the action of $T$, $M$ is a locally nilpotent module, every finite quotient is a nilpotent module, but since $T$ is surjective, this means that the only finite quotient module of $M$ is reduced to $\{0\}$. So $M$ is not a residually finite module (for $T$, and hence for $U$) and thus $G$ is not residually finite.
(3) Here's a counterexample, with $N$ finite. Fix a prime $p$ and fix a non-abelian group $F$ of order $p^3$, with two non-commuting elements $x,y$. Let $F_n$ be a copy of $F$, with the corresponding elements $x_n,y_n$, and $z_n=[x_n,y_n]$. Let $G$ be quotient of the direct sum $\bigoplus_n F_n$ by identifying all central elements $z_n$ (thus defining a single central element $1\neq z=[x_n,y_n]$ of order $p$). Then $G/\langle z\rangle$ is abelian and residually finite. But $G$ is not residually finite: in a finite quotient, there exist $n\neq m$ such that $x_n$ and $x_m$ have the same image. Hence $z=[x_n,y_n]$ and $[x_m,y_n]=1$ have the same image. Hence $z$ is killed in every finite quotient.
(4) For a split extension, if $N$ is finitely generated, it is easy to see that $G$ is residually finite (without the abelian assumption).
(5) If $G$ itself is finitely generated, it is an observation of Ph. Hall that indeed $G$ is residually finite (as mentioned in Anthony's answer). Indeed, if $Q$ is finitely generated abelian, then $G$ is residually finite if and only if $N$ is a residually finite $Q$-module, and this indeed holds when $N$ is a finitely generated $Q$-module, but not in general as (2) above shows (even when the underlying group $N$ is residually finite).