I will focus on reductive Lie algebras $\mathfrak{g}$ of semisimple rank $\geq 1$ (so they're not abelian, for which the question is easy).
In that case, the set of semisimple element is dense in $\mathfrak{g}$, but is neither open, nor closed, nor locally closed in the Zariski topology.
Density is easy to see in $\mathfrak{sl}_n$, because any matrix whose characteristic polynomial has no multiple roots over $\bar{k}$ (such matrices form a non-empty open hence dense subset) is semisimple. In general, the same proof works, noting that the set of regular semisimple elements is a dense open subset of $\mathfrak{g}$.
The set of nilpotent elements $\mathcal{N}$ in $\mathfrak{g}$ is Zariski closed (this being defined by the vanishing of all polynomial invariants of the Lie algebra); in the case of $\mathfrak{sl}_n$ this is because nilpotency is defined by having characteristic polynomial coefficients equal to zero.
If $\mathfrak{g}_s$ were open in $\mathfrak{g}$, the set of semisimple nilpotent elements $\mathfrak{g}_s\cap \mathcal{N}$ would be open in $\mathcal{N}$. However, the only element that is both nilpotent and semisimple is zero, and $\{0\}$ is not open in $\mathcal{N}$. (The last sentences uses the fact that the semisimple rank is $\geq 1$, so $\mathcal{N} \neq \{0\}$ because it contains for example regular nilpotent elements.)
It's now easy to see that $\mathfrak{g}_s$ can't be closed or even locally closed either. Indeed, if it would be, it would be open in its closure. But we showed that its closure is $\mathfrak{g}$ and we showed it's not open in $\mathfrak{g}$!
If you're looking for a positive result, the best thing I can think of is the following: let $G$ be a reductive group with Lie algebra $\mathfrak{g}$ and write $S = \mathfrak{g} // G$ for the space of invariant polynomials for the adjoint action of $G$ on $\mathfrak{g}$. Let $\pi \colon \mathfrak{g} \rightarrow \mathfrak{g} // G$ be the morphism of taking invariants. In the case of $G = SL_n$, $\mathfrak{g} // G$ is the space of monic polynomials with no $x^{n-1}$ term and $\pi$ is the morphism of sending a matrix to its characteristic polynomial. Then the theorem is (I think this is due to Kostant) that the intersection of $\mathfrak{g}_s$ with every fibre of $\pi$ is closed. (Taking the fibre above zero, you get the subvariety of nilpotent elements.) So somehow $\mathfrak{g}_s$ is well-behaved and closed in every fibre where you have fixed the invariants, but packaging all the fibres together gives a slightly weird set with bad topological properties.
If you want to read more about this, I highly recommend Kostant's papers 'Lie Group Representations on Polynomial Rings' and 'The Principal Three-Dimensional Subgroup and the Betti Numbers of a Complex Simple Lie Group'.
Best Answer
Yes, and a more general statement is true even over general fields of characteristic $0$, according to Bourbaki's take on Cartan subalgebras (in book VII, ยง2 of the volume on Lie Groups and Lie Algebras). Namely, proposition 10 says that for an abelian subalgebra $\mathfrak{a} \subseteq \mathfrak{g}$ consisting of semisimple elements,
$$\lbrace \text{Cartan subalgebras of } \mathfrak{g} \text{ containing } \mathfrak{a} \rbrace = \lbrace \text{Cartan subalgebras of } \mathfrak{z}_{\mathfrak{g}}(\mathfrak{a}) \rbrace $$
($\mathfrak{z}_{\mathfrak g} =$ centraliser). But every Lie algebra has Cartan subalgebras (see e.g. corollary 1 to theorem 1 loc. cit.), in particular so does $\mathfrak{z}_{\mathfrak{g}}(\mathbb{C} J)$ in your question.