This is true and follows from:

**Claim:** Let $x$ be a $n\times n$ matrix with $\mathbb{C}$-coefficients. Then the centralizer $C(x)$ of $x$ in $GL_n(\mathbb{C})$ fits into a short exact sequence $1\rightarrow U\rightarrow C(x) \rightarrow \prod_{n_i} GL_{n_i}\rightarrow 1$, where $U$ is a unipotent group and $n_i$ a sequence of integers.

Since an extension of special groups is special and since $\mathbb{G}_a$ and $GL_k$ are special, $C(x)$ is special too, answering your question. For the proof of the claim:

- It suffices to prove the claim for $x$ nilpotent. Indeed, let $x = x_s+x_n$ be the Jordan decomposition into semisimple and nilpotent parts. Then $C(x)$ is the centralizer of $x_n$ in $C(x_s)$. Since $C(x_s)$ is a product of general linear groups (easy to see by taking $x_s$ a diagonal matrix) and $x_n$ is a nilpotent element of $Lie(C(x_s))$ (a product of matrix Lie algebras), we are done by looking at each factor.
- For $x$ nilpotent, $C(x)$ is an extension of a unipotent group and a product of general linear groups. This follows from the explicit computation in section IV.1.7-1.8 in the paper below.

*Springer, T. A.; Steinberg, R.*, Conjugacy classes, Sem. algebr. Groups related finite Groups Princeton 1968/69, Lect. Notes Math. 131, E1-E100 (1970). ZBL0249.20024.