Center of a non-abelian subgroup of $GL(2, \mathbb{C})$

First, suppose that $J$ is diagonal but not scalar. That is, $J=\begin{pmatrix}a&0\\0&b\end{pmatrix}$ with $a\neq b$. Let $X=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ be an element of $G$. Then, $XJ=JX$ yields $$\begin{pmatrix}ap&bq\\ar&bs\end{pmatrix}=\begin{pmatrix}ap&aq\\br&bs\end{pmatrix}.$$ That is, $q=0$ and $r=0$. Thus, every matrix in $G$ is diagonal, and so $G$ is abelian.

Finally, we suppose that $J$ is a nontrivial Jordan block. Then, $J=\begin{pmatrix}k&1\\0&k\end{pmatrix}=kI+E$ with $k\neq 0$, where $E=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. Again, let $X=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ be an element of $G$. Then, $XJ=JX$ yields $$\begin{pmatrix}kp&p+kq\\kr&r+ks\end{pmatrix}=\begin{pmatrix}kp+r&kq+s\\kr&ks\end{pmatrix}.$$ This shows that $r=0$ and $p=s$, so $X=pI+qE$. Hence, $G$ contains only polynomials in $E$, and they all commute.