Under what conditions does the second cohomology preserve injectivity?

Sasha's comment is correct. In your case you have the trivial action on $p$-power order cyclic modules. So let me write the map $B to B/A$ as $B = mathbb Z/p^n to
mathbb Z/p^m = B/A,$ where $m leq n$, and the map is the natural one (reduce a mod $p^n$ class to a mod $p^m$ class).

Now $H^1$ of $G$ against a trivial module is just homs of $G^{ab}$ into this module, so
we have to look at
$$Hom(G^{ab},mathbb Z/p^n ) to Hom(G^{ab},mathbb Z/p^m).$$
Now $G^{ab}$ is (in your setting) itself an abelian $p$-power order group, so is a product
of cyclic $p$-power order groups, so (since $Hom$ from a finite product is the product
of the individual $Hom$s) we see that it is enough to consider whether
$$Hom(mathbb Z/p^r,mathbb Z/p^n) to Hom(mathbb Z/p^r,mathbb Z/p^m)$$ is surjective.

Assuming that $m < n$ (i.e., in the original terms of the problem, that $A$ is non-zero,
so that the question is non-trivial), then this map is surjective if and only
if $n leq r.$ (This is not hard to check; see below for a careful explanation.)

Putting this together for all $r$, we get the following: assuming that $A$ is non-zero, that $B$
is cyclic of $p$-power order, and that $G$ is a $p$-group, then the map of $H^2$ is injective precisely when each cyclic direct summand of $G^{ab}$ has order at least
that of $B$.

Proof of surjectivity fact:
The $Hom$ space $Hom(mathbb Z/p^r,M)$ is equal to $M[p^r],$ the $p^r$-torsion subgroup of $M$,
for any abelian group $M$ (just look at the image of $1$ mod $p^r$). So we have to consider
the surjectivity (or non-surjectivity) of
$(mathbb Z/p^n)[p^r] to (mathbb Z/p^m)[p^r],$ which
is the map
$$p^{max(0,n-r)}mathbb Z/p^nmathbb Z to p^{max(0,m-r)}mathbb Z/p^n mathbb Z.$$
This is surjective if $n - r leq 0,$ or if $m = n$, but otherwise is not.