Let $z_0\in\Bbb{R}$ be arbitrary. The matrices $A_0:=A(z_0)$ and $B_0:=B(z_0)$ are normal; in view of $A_0^*=A^\#(\bar{z_0})=A^\#(z_0)$ and $B_0^*=B^\#(\bar{z_0})=B^\#(z_0)$ they commute with their Hermitian adjoints. They also commute with each other. So $A_0$ and $B_0$ could be simultaneously diagonalized by a unitary matrix. That matrix also diagonalizes $A_0^*$ and $B_0^*$. So the four matrices $A_0=A(z_0),B_0=B(z_0),A_0^*=A^\#(z_0),B_0^*=B^\#(z_0)$ could be simultaneously diagonalized which implies that they all commute with each other. So $B(AA^\#)=(AA^\#)B$ at any point of the real line. Because the entries are entire functions, they coincide throughout the whole complex plane.
Added: As mentioned by @Guest and @MarkSapir, $A^\#(z)$ (respectively $B^\#(z)$) is the Hermitian conjugate of $A(\bar{z})$ (resp. $B(\bar{z})$) rather than that of $A(z)$ (resp. $B(z)$). Indeed, the Hermitian conjugates $A^*(z):=(\overline{A(z)})^{\rm{T}}$ and $B^*(z):=(\overline{B(z)})^{\rm{T}}$ do not vary holomorphically with $z$; they are anti-holomorphic. On the contrary, $A^\#(z)$ and $B^\#(z)$ are holomorphic because in their definitions the complex conjugation is applied both in the domain and the range. To circumvent this difficulty, one notices that $A^\#(z)$ (respectively $B^\#(z)$) is indeed the Hermitian conjugate of $A(z)$ (resp. $B(z)$) if $z$ is real. So by a linear algebra argument, one can establish $B(z)\left(A(z)A^\#(z)\right)=\left(A(z)A^\#(z)\right)B(z)$ for $z\in\Bbb{R}$. This persists on the rest of $\Bbb{C}$ because the entries of these matrices are entire functions (holomorphic on the whole $\Bbb{C}$).