I'm not sure whether this qualifies as "elementary", but here's one way:

Note that $A$ and $B$ form a commuting family of matrices, so that $A$ and $B$ are simultaneously upper-triangularizable. That is, there is some unitary $U$ such that $U^*AU$ and $U^*BU$ are both upper triangular.

Note that $U^*AU$ and $U^*B U$ are both normal and upper-triangular, which means that they are diagonal. That is, we have
$$
A = U\pmatrix{
\lambda_1 & &\\
&\ddots&\\
&&\lambda_n}U^* \quad
B = U\pmatrix{
\mu_1 & &\\
&\ddots&\\
&&\mu_n}U^*
$$
From here, it is easy to verify that $AB^* = B^*A$.

Alternatively, there's a nice inductive proof here if we use the following two facts:

- Commuting matrices have a common eigenvector
- A matrix of the form $\pmatrix{A&B\\0&C}$ is only normal if $B = 0$

With that, we may use the fact that $A$ and $B$ have a common eigenvector to write
$$
A = U\pmatrix{
\lambda_1 & c^T\\
0&\tilde A}
U^* \quad
B = U\pmatrix{
\mu_1 & d^T\\
0&\tilde B}U^*
$$
for a unitary $U$ (whose first column is a common eigenvector to $A$ and $B$). Because $A$ and $B$ are normal, we see that $c = d = 0$. Because $A$ and $B$ are commuting normal matrices, $\tilde A$ and $\tilde B$ are (smaller) commuting normal matrices.

By induction, we may conclude (as before) that for some unitary $U$, we have
$$
A = U\pmatrix{
\lambda_1 & &\\
&\ddots&\\
&&\lambda_n}U^* \quad
B = U\pmatrix{
\mu_1 & &\\
&\ddots&\\
&&\mu_n}U^*
$$
and we may reach the desired conclusion, just like last time.