See Kuenneth-formula for group cohomology with nontrivial action on the coefficient
Let $C=C_2$ be the cyclic group of order two, $\def\ZZ{\mathbb Z}\ZZ$ the trivial module over $C$ and $S$ the $C$-module which is $\ZZ$ as an abelian group with with the non-trivial action of $C$. You want to compute $H^\bullet(C\times C,M)$ with $M=\ZZ\otimes S$, and the Künneth formula tells us that there is a short exact sequence $$0\to (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^p\to H^p(C\times C,M)\to \bigoplus_{r+s=p-1}Tor^\ZZ(H^r(C,\ZZ),H^s(C,S))\to0$$
You can easily compute $H^\bullet(C,-)$ with coefficients in every $C$-module and the Tor is also easy to compute. If I am not making too many mistakes, then \begin{gather} (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=0 \end{gather} for all even $p\geq0$, and \begin{gather} (H^\bullet(C,\ZZ)\otimes H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^r,\\ (Tor_1(H^\bullet(C,\ZZ),H^\bullet(C,S))^{p}=(\ZZ/2\ZZ)^{r-1} \end{gather} for all odd $p=2r-1\geq0$. If follows that in the short exact sequenc above exactly one of the first and third terms are not zero for a given $p$, so we get an isomorphism in all cases.