The answer is no in general. For instance, by what is essentially an argument of Sierpiński, if $(X,\Sigma,\nu)$ is a $\sigma$-finite continuous measure space, then no non-null subset of $X$ admits a $\nu\times\nu$-measurable well-ordering. The proof is almost verbatim the one here.
It is consistent (assuming large cardinals) that there is an extension of Lebesgue measure defined on all sets of reals. Here, $X=\mathbb R$ and $\Sigma=\mathcal P(\mathbb R)$. Since $\nu$ extends Lebesgue measure, the space satisfies the assumptions of the result just stated, and $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is not $\mathcal P(\mathbb R^2)$.