Are the weak* and the sequential weak* closures the same?

The answer is no.

For a specific, non-separable counter-example take $E=\ell_\infty$. Then the set $$X=\{e_n^*\colon n\in \mathbb{N}\}\subseteq \ell_\infty^*$$ consisting of evaluation functionals is bounded but not weak* compact. (Here $\langle e_n^*, f\rangle = f(n)$ for $f\in \ell_\infty$.) However, by boundedness, the weak* closure of $X$ is compact in the weak* topology hence different from $X$. There are no non-eventually constant convergent sequences in $X$ so the cluster points of $X$ cannot be realised as limits of sequences.

To be more precise, the weak*-sequential closure of $X$ is $X$ itself but $X$ is not weak*-closed (because it is not weak*-compact) so the weak* closure of $X$ is strictly bigger than $X$.