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](https://math.stackexchange.com/a/88811/462).

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 $\mathcal P(\mathbb R^2)$.

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By the way, there is a recent article in the *Monthly* dealing precisely with this problem and discussing how $\mathsf{CH}$ implies that $\mathcal P(\mathbb R)\otimes\mathcal P(\mathbb R)$ is $\mathcal P(\mathbb R^2)$ while the existence of extensions of Lebesgue measure gives a negative answer:
> [MR3626256][1] Avilés, Antonio; Plebanek, Grzegorz. [*A little ado about rectangles*][2]. Amer. Math. Monthly **124** (2017), no. 4, 345–350. 


  [1]: http://www.ams.org/mathscinet-getitem?mr=3626256
  [2]: http://www.jstor.org/stable/10.4169/amer.math.monthly.124.4.345