Let $(e, h, f)$ be an $\operatorname{SL}_2$-triple in $\mathfrak g$. My understanding is that $e + C_{\mathfrak g}(f)$ is called a Kostant section only in case $e$ is regular; but I don't impose this restriction. (EDIT: It seems that, in general, it's called a Slodowy slice.) Given such a datum, there is a unique decomposition (*) $\mathfrak g = [e, \mathfrak g] + C_{\mathfrak g}(e, f) + C_{\mathfrak g}(f)(< 0)$, where “$< 0$” in the last term refers to the grading coming from the action of $h$.

Now suppose that we are given only a nilpotent element $e \in \mathfrak g$, and an arbitrary element $X \in \mathfrak g$. Can we complete $e$ to an $\operatorname{sl}_2$-triple $(e, h, f)$ so that, in the decomposition $X_1 + X_2 + X_3$ of $X$ according to (*), we have that $X_2$, or at least its semisimple part, commutes with $X_3$? The obvious answer is “No, because, in certain characteristic, you can't complete $e$ to an $\operatorname{sl}_2$-triple at all.” To avoid such trivialities, assume the characteristic is as large as necessary.

As a very mild hint that the answer is ‘yes’, the weak form of the question works whenever $e$ is distinguished (since then the semisimple part of $X_2$ is central in $\mathfrak g$); and, by explicit calculation, it works when $\mathfrak g = \mathfrak{gl}_3$ and $e$ is the $(1, 2)$-nilpotent, as long as $p > 3$. Because of the first hint, I'd like to make an appeal to Bala–Carter(–Pommerening–Premet …) theory, but I can't figure out how.