Let $A=\left\{\begin{pmatrix} a & b \\ -\overline{b} & \overline{a}\end{pmatrix}\ \Big\vert\,\, a,b\in\mathbb{C}\right\}$. I want to determine $Z(A)$.
I've seen that $A$ is actually the matrix representation of the qauternions, so what I've found by searching the web is that $Z(A)=\{\alpha I\ |\ \alpha\in\mathbb{R}\}$. However I'm trying to calculate $A$ explicitely.
So if $x\in Z(A)$ then $xy=yx$ for all $y\in A$. So by taking $y=\begin{pmatrix} i & 0 \\ 0 & -i\end{pmatrix}\in A$ then for $x=\begin{pmatrix} a & b \\ -\overline{b} & \overline{a}\end{pmatrix}$ we should have
$$\begin{pmatrix} ia & -ib \\ -i\overline{b} & -i\overline{a}\end{pmatrix}=\begin{pmatrix} ia & ib \\ -i\overline{b} & -i\overline{a}\end{pmatrix}$$
So we may conclude $b=0$ as $-ib$ should equal $ib$. Likewise if we take $y=\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix}\in A$ we should have
$$\begin{pmatrix} b & a \\ -\overline{a} & -\overline{b}\end{pmatrix}=\begin{pmatrix} -\overline{b} & \overline{a} \\ -a & -b\end{pmatrix}$$
So $a$ should be real as $a=\overline{a}$.
Does this proof work?
Yes, your strategy is correct. Technically you should add a comment at the end that all such matrices are obviously elements of the centre because they're proportional to the identity matrix, but yes, one can constrain centre elements by identifying specific matrices that are hard to commute with.