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@NateEldredge Do you see now why I got stuck? Because this statement is not true, see a very suprising counterexample by R W below! Do you still think that it was "a straightforward exercise in the monotone class"?
@NateEldredge It's not so clear to me how the monotone class lemma would work here. Let's say we proved that if an invariant set $A$ from $E\times E$ (I denoted by $E$ the state space of the Markov process) is of the form $A_1\times A_2$ then $\mu\otimes\mu (A)$ is either $0$ or $1$. Such sets definetely form a $\pi$-system, let's call it $\mathcal{A}$. It's clear that all invariant sets form a $\sigma$-algebra (and hence a $\lambda$-system), let's call it $\mathcal I$. By the monotone class theorem we have $\sigma(\mathcal{A})\subset \mathcal I$. But why $\sigma(\mathcal{A})=\mathcal I$?
