Let $H$ be the centralizer of $a_0$.  $H$ certainly contains the centralizer $MA$ of $A$.  We now compare their Lie algebras, $\mathfrak{h}$  and $\mathfrak{m}+\mathfrak{a}$.  Since $H$ contains $A$ it is normalized by it and we can decompose $\frak{h}$ under this action.  Now $\frak{m}+\frak{a}=\frak{g}_0\subset\frak{h}$ is the subspace corresponding to the trivial character of $A$, so the rest of $\frak{h}$ must transform under other characters (that is, under the roots).  However, no root vanishes on $a_0$ (since $a_0$ is not on any wall), so no root can occur in this action.  It follows that $\frak{h} = \frak{a}+\frak{m}$.

This shows that the connected components of $H$ and $MA$ agree.

In the algebraic category, the centralizer of $A$ is Zariski-connected, and $A$ is the maximal split torus in the center of the centralizer, hence characteristic.  It follows that $A$ is normal in $H$, so $H$ is contained in the normalizer of $A$.  But $N_G(A) / Z_G(A)$ is the Weyl group, which acts simply transitively on the chambers.  It follows that $H = Z_G(A)$.

In the smooth category, $\frak{a}$ is the unique Cartan subalgebra of $MA$, so again $A$ is characteristic there and $H \subset N_G(A)$, at which point the argument can proceed in the same fashion.

[Edited: erroneous claim about connectedness of $H$ replaced with correct discussion.  Reference added]

[Edited: I think I have a reference: Lemma 6.4.3 in Springer's "[Linear Algebraic Groups][1]", 2nd edition.  The result is not exactly the same (it concerns regular elements of $A$, not of $\frak{a}$), and the arguments depend
on an embedding in $\mathrm{GL}_n$, but it may be enough for you.]

   [1]: http://link.springer.com/book/10.1007%2F978-0-8176-4840-4 "Springer: Linear Algebraic Groups, 2nd edition"