After thinking about this for a little while, it seems like there is a satisfactory way of carrying this out in group cohomology using the Hochschild-Serre spectral sequence of a group extension
$$1 \to H \to G \to Q \to 1$$
(henceforth we will call this the HSSS), and since my original motivating situation happened to be a fibration of aspherical manifolds, this will suffice for my purposes. I will refrain from selecting this as the answer, though, since I would still be curious to know how to carry this out in the topological setting.
For simplicity, I'll restrict attention to the trivial coefficient module $\mathbb Z$. Suppose that $H$ is a $PD_{m}$-group, $Q$ is a $PD_{n}$-group, and $G$ is a $PD_{m+n}$-group. Chapter VII of Brown's Cohomology of Groups contains a construction of the HSSS. In the homological setting, the idea is to show that there is a particular chain complex $C$ for which the homology with coefficients in the chain complex $H_*(Q, C)$ is isomorphic to $H_*(G)$; the HSSS is then derived from one of the filtrations on the bigraded chain complex underlying the definition of $H_*(Q, C)$.
In fact, letting $F$ be a projective resolution of $\mathbb Z$ over $\mathbb Z G$ (which can be assumed to be of length $m+n$), Brown explains why $C = F_H$ suffices. If we let $F'$ be a projective resolution of $\mathbb Z$ over $\mathbb Z Q$, the upshot is that the tensor product of chain complexes $F' \otimes_Q F_H$ computes $H_*(G)$.
Turning to cohomology, similar arguments show that the homology of the cochain complex
$$
\operatorname{Hom}_Q(F', \operatorname{Hom}_H(F, \mathbb Z))
$$
is given by $H^*(G)$.
The crucial observation is that for a $PD_k$-group $\Gamma$, if $F$ is a projective resolution of finite length $k$, then $\overline F : = \operatorname{Hom}_\Gamma(F, \mathbb Z \Gamma)$ is also a projective resolution of $\mathbb Z$ over $\mathbb Z \Gamma$. (A technical point which explains how the shift in degrees arises is that one must monkey with the grading on $\overline F$ in the proper way).
We can apply tensor-Hom adjunction to see that
$$
(*)\quad \operatorname{Hom}_Q(F', \operatorname{Hom}_H(F,\mathbb Z)) \cong \overline{F'} \otimes_Q \operatorname{Hom}_H(F,\mathbb Z) \cong \overline{F'} \otimes_Q (\operatorname{Hom}_H(F,\mathbb Z H) \otimes_H \mathbb Z).
$$
As $Q$ is a $PD$-group by assumption, we may use $\overline{F'}$ as our projective resolution in the computation of $H_*(Q,F_H)$. I next claim that after an adjustment to the grading, (a portion of) the chain complex $\operatorname{Hom}_H(F,\mathbb Z)$ is weakly equivalent to $F_H$. This follows from the fact that $H$ is also a $PD$-group. More specifically, the assumption that $H$ is a $PD_m$ group is equivalent to the condition that $H^i(H, \mathbb Z H)$ vanish unless $i = m$, in which case it should be infinite cyclic. This says that the portion
$$
\operatorname{Hom}_H(F_0, \mathbb Z H) \to \dots \to \operatorname{Hom}_H(F_{m-1}, \mathbb Z H) \to \operatorname{Hom}_H(F_{m}, \mathbb Z H)
$$
of the chain complex $\operatorname{Hom}_H(F, \mathbb Z H)$ forms a projective resolution of $\mathbb Z$ over $\mathbb Z H$, from which the claim follows.
Paying attention to the bigradings, the $(p,q)$-component of the left-hand side in $(*)$ corresponds to the $(n - p, m-q)$-component on the right.
Now the properties I was looking for above will follow automatically from the spectral sequence machinery.
