I can offer the following generalization for non-trivial coefficients $A$: Write $Q = G/H$. The cap product $$H_1(H;\mathbb{Z}) \otimes H^1(H;A) \to \mathbb{Z}\otimes_H A = A_H$$ is $G$-linear with trivial $H$-action and induces a cup product
$$\cup: H^p(Q;H_1(H;\mathbb{Z})) \otimes H^q(Q;H^1(H;A))\xrightarrow{} H^{p+q}(Q;A_H).$$
Let $u \in H^2(Q;H_1(H;\mathbb{Z}))$ be the class that corresponds to the extension
$$1 \to H^{ab} \to G/H' \to Q \to 1$$
and let $\kappa: A^H \to A_H$ be the canonical homomorphism.

**Theorem:** The compostion
$$H^p(Q;H^1(H;A)) \xrightarrow{d_2}H^{p+2}(Q;A^H)\xrightarrow{\kappa^\ast}H^{p+2}(Q;A_H)$$
is (up to sign) cup product with $u$, i.e. $(\kappa^\ast \circ d_2^{p,1})(x) = - u \cup x$.

*Proof:* By abuse of notation let $\kappa: A \to A_H$. Since $\kappa$ is $G$-linear, it induces a map of spectral sequences $\kappa_r^{pq}: E_r^{pq}(A) \to E_r^{pq}(A_H)$. In particular, $\kappa_2^{p+2,0}\circ d_2^{p,1}(A)=d_2^{p,1}(A_H)\circ \kappa_2^{p,1}$. Clearly, $\kappa_2^{p+2,0}$ is the map $\kappa^\ast$ in the theorem and since $H$ acts trivially on $A_H$, we obtain $(\kappa^\ast \circ d_2^{p,1}(A))(x) = d_2^{p,1}(A_H)(\kappa_2^{p,1}x)=-u \cup \kappa_2^{p,1}x$. Hence, it remains to show
$u \cup x = u \cup \kappa_2^{p,1}x$ (note the cup products are w.r.t. different pairings). But this follows immediately by applying $H^\ast(Q;-)$ to the following commutative diagram of pairings:
$$\begin{array}{ccc}
H_1(H;\mathbb{Z}) \otimes H^1(H;A) \;\; & \xrightarrow{\cap} & A_H \newline
\scriptstyle id\otimes \kappa^\ast \displaystyle\downarrow\qquad\quad & & \downarrow\scriptstyle id \newline
H_1(H;\mathbb{Z}) \otimes H^1(H;A_H) & \xrightarrow{\cap} & A_H.
\end{array}$$