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]

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", 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.]

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 we are done (the centralizers of a semisimple element and of a torus are both connected). I can't recall the argument in the smooth category off the top of my head.

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]