When $A$ acts trivially on $N$, this $H^2(A,N)$ is, if I'm correct, naturally isomorphic to the group $\mathrm{Hom}(\Lambda^2A,N)$ ($\Lambda^2A$ being the second exterior power, quotient of $A\otimes_\mathbf{Z}A$ by the subgroup generated by elements of the form $x\otimes x$ when $x$ ranges over $A$), and the corresponding extension has $N$ as derived subgroup iff the corresponding element $f\in\mathrm{Hom}(\Lambda^2A,N)$ is surjective.

In general, let $N/M$ be the co-invariants of the $A$-action on $N$ (i.e., $M$ is generated as a group by the $\varphi(g)h-h$, when $h$ ranges over $N$ and $g$ over $A$). Then after modding out by $M$, every extension as given yields an $f\in\mathrm{Hom}(\Lambda^2A,N/M)\simeq H^2(A,N/M)$; then $N$ is the derived subgroup iff $f$ is surjective.

When $A$ acts trivially on $N$, this $H^2(A,N)$ has a canonical map $\Phi$ into the group $\mathrm{Hom}(\Lambda^2A,N)$ ($\Lambda^2A$ being the second exterior power, quotient of $A\otimes_\mathbf{Z}A$ by the subgroup generated by elements of the form $x\otimes x$ when $x$ ranges over $A$), induced by the commutator map. Given a central extension with cocycle $c$, the resulting extension has $N$ as derived subgroup iff the corresponding element $\Phi(c)\in\mathrm{Hom}(\Lambda^2A,N)$ is surjective.

(Remark: there's a canonical inclusion $\Psi$ of $\mathrm{Hom}(\Lambda^2A,N)$ into $H^2(A,N)$ and $\Phi\circ\Psi=2\mathrm{Id}$.)

In general, let $N/M$ be the co-invariants of the $A$-action on $N$ (i.e., $M$ is generated as a group by the $\varphi(g)h-h$, when $h$ ranges over $N$ and $g$ over $A$). Then after modding out by $M$, every extension as given yields a cocycle in $H^2(A,N/M)$ and hence, taking $\Phi$ an element in $f\in\mathrm{Hom}(\Lambda^2A,N/M)$; then $N$ is the derived subgroup iff $f$ is surjective.