This question concerns the (co)homology of groups.
How well does the functor $H^\ast(G;-)$ play with tensor products of $G$-modules? Are there nice general statements one can make when one's coefficients are of the form $M\otimes N$?
Any references? (I can't find anything to this end in Brown)
It has been a long time since you asked that and you have probably sorted it out all alone, however since it is still open and someone else might me interested in, I shall give it a shot (of course as I said if you have figured it out feel free to add your idea/understanding behind this).
To start off, unfortunately there is definitely not good interaction between the functors ${\rm H}^{*}(G,-)$ and $-\otimes-$ (well the later is a bifunctor strictly speaking, but let's become a bit sloppy for a while); as Qiaochu commented above the problem starts early on, that is, for $n=0$. Remember that ${\rm H}^0(G,M)=M^G$, $\forall M \in \mathsf{mod}_G$. In fact by definition group cohomology is defined to be the right derived functor of,$\,\,$ $-^G: \mathsf{mod}_G \to \mathsf{mod}_{\mathbb{Z}}$.
Remark: The above functor is right adjoint to the trivial module functor $\mathsf{mod}_{\mathbb{Z}} \to \mathsf{mod}_G$ (exercise), fact that proves the left exactness immediately (although it is kind of obvious anyway).
Instead of giving you an example (Qiaochu's example is excellent and simple to digest it), I will give you a more general reason why we can't hope to have a good relation in the following lemma.
Lemma: Assume $k$ is an algebraically closed field of positive characteristic and $G$ a finite group. If $V$ is an arbitrary $kG$-module and $P$ is a projective $kG$-module, then the tensor product $V \otimes P$ is projective too.
In view of the above lemma, and since equivalently we can define the group cohomology groups as ${\rm H}^n(G;M) \cong {\rm Ext}^n_{kG}(k,M)$, projectiveness of a module $M$ implies trivial cohomology groups (note that $\mathsf{mod}_{kG}$ is a Frobenius category). Hence if we choose a module $V$ with non-trivial, say $n$-cohomology, then the tensor product with $P$ forgets the cohomological structure of $V$ since it "kills" it. I wouldn't say that this is a pathology of $\mathsf{mod}_{kG}$, but rather a useful for other purposes intrinsic property of this category. However in my eyes makes some sort of obstruction towards the direction we are looking at this moment.
However, because it's good to illustrate a particular example where some kind of behaviour exists think the following. Choose $G=\mathbf{C}_p$, the cyclic group or order $p$ and a field $k$ as in the lemma above. Then for a given $n$-dimensional representation over $k$, say $\rho_1 : G \to {\rm GL}(W)$, we can always choose a triangulated equivalent version of $\rho_1$, that is $\rho_2 : G \to {\rm UT}(V)$, (the latter means the upper-triangular matrices with $1_{k}$ along the diagonal), while choosing a basis for the dual $W^*$, $\{ x_1,...,x_n \}$, we can think of the symmetric algebra ${\rm Sym}(W^*)$ as the polynomial algebra in $n$- variables (there exists a canonical graded isomorphism of $k$-algebras), $k[x_1,...,x_n]:=k[W]$. So someone naturally can think of $k[W]$ as a $kG$-module via $\rho_2$, which acts by degree-preserving algebra automorphisms on $k[W]$. A question now arises, what can be said about ${\rm H}^n(G;k[W])$? This question is generally really hard and goes deep, but for this case there is a convincing answer. Firstly, notice that we can form an hsop $B=k[{\rm \bf{N}}_G(x_1),{\rm \bf{N}}_G(x_2),...,{\rm \bf{N}}_G(x_n)]$, where $N_i:={\rm \bf{N}}_G(x_i)= \prod_{j=1}^n \sigma^j(x_i)=x^p_i$, are the so-called norm elements and that $ B \subset k[W]$ is a Noether's Normalization, since $\{ {\rm \bf{N}}_G(x_1),{\rm \bf{N}}_G(x_2),...,{\rm \bf{N}}_G(x_n) \}$ forms a Gröbner Basis. Because of the latter in particular, for any $f \in k[W]$ we can form the so-called norm decomposition, $f=f_1N_1+f_2N_2+...+f_nN_n+r$. So we can define the $\mathbf{C}_p$-submodule $k[W]^{\flat}$ of polynomials where $f_i=0, \forall i=1,...,n$. Now from all the above should be kind of straightforward that we have a $B$-module isomorphism (bewteen free $B$-modules) $$k[W] \cong B\otimes_{k}k[W]^{\flat}.$$
Proposition: ${\rm H}^n(\mathbf{C}_p,k[W]) \cong {\rm H}^n(\mathbf{C}_p,B\otimes_{k}k[W]^{\flat}) \cong B \otimes_k {\rm H}^n(\mathbf{C}_p;k[W]^{\flat})$, for $n>0$.
So the cohomology groups in that case behave in some sense naturally. Of course this is not the case is general, and I presented this example just to delineate the complexity of the problem when the situation is not ideal (the group is not cyclic, the underlying ring is not field is not algebraically closed and others...). For any possible questions please do let me know (if you are still in this community :)).
P.S.1. The proof of the above is omitted since it is not what you care about at the moment I guess, but it comes from Universal Coefficient Theorem.
P.S.2. Generally if you want to check further interaction between ${\rm H}^r(G; A), {\rm H}^s(G; B)$ and ${\rm H}^{r+s}(G; A \otimes B)$, check the idea of cup products.