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 gorup an uncountable group in which every proper subgroup is countable). (Sciencedirect link to Shelah's 1980 article)

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)