The inclusion can indeed be proper. Indeed, there exist simple groups $S$ with no maximal subgroup. So $\Phi(S)=S$. But the diagonal in $S\times S$ is a maximal subgroup, and in particular contains $\Phi(S\times S)$. Since $\Phi(S\times S)$ is normal, it follows that $\Phi(S\times S)=\{1\}$, which is properly contained in $\Phi(S)\times\Phi(S)=S\times S$.
I'm not sure what's the simplest example of such $S$, but at least one comes from Shelah's construction of a simple Jonsson group (a Jonsson group an uncountable group in which every proper subgroup is countable). (Sciencedirect link to Shelah's 1980 article)
(added: there also exists a simple countable group without maximal subgroups: Theorem 35.3 in Olshanskii's book Geometry of defining relations in groups)