First take the case $n=1$. Let $A(z)=(a(z))$, $B(z)=(b(z))$, $a(z),b(z)$ entire functions. Then $A(z)A^*(z)+B(z)B^*(z)=|a(z)|^2+|b(z)|^2$. Therefore we need $C(z)=(c(z))$ with $c(z)$ an entire function s.t $A(z)A^*(z)+B(z)B^*(z)=C(z)C^*(z)=|C(z)|^2=|a(z)|^2+|b(z)|^2$ for all $z\in \mathbb {C}$.
Therefore $|C(z)|^2\geq |a(z)|^2$ and $|C(z)|^2\geq |b(z)|^2$ for all $z\in \mathbb {C}$.
Liouville's theorem says that no entire function dominates another unless they are related by a constant multiple. Hence we must have $a(z)=\lambda b(z)$for some $\lambda \in \mathbb {C}$ and then $c(z)=\sqrt{1+|\lambda|^2}b(z)$.
Thus for $n=1$ we require that $A$ and $B$ are scalar multiples of one another and this also satisfies the more general equation with $z$ and $w$.
I believe in general you may need $A(z)=T(z)M$ and $B(z)=T(z)N$ where $M, N \in \mathbb{C}_{n\times n}$, constant matrices and $T(z)$ is any $n \times n$ matrix with entire entries.
This works because then
$$A(z)A^*(w)+B(z)B^*(w)=T(z)(MM^*+NN^*)T^*(w)$$$$=(T(z)Q)(T(w)Q)^*=C(z)C(w)^*$$
where $C(z)=T(z)Q$ a matrix with entire entries and $QQ^*=MM^*+NN^*$. $Q$ exists because $MM^*+NN^*$ is hermitian and has a Cholesky Decomposition.
To prove this we need the following lemma:
Lemma: For any $n \times n$ matrix $A(z)$ with entire entries there exists a matrix $A'(z)$ also with entire entries such that $A'(z)A(z)=A(z)A'(z)=D(z)I$ where $D(z)$ is entire.
Proof: $A'$ is just the Adjugate Matrix of $A$ and $D(z)=\det(A).$ $\blacksquare$
Theorem:
If $A(z)$, $B(z)$ and $C(z)$, all $n \times n$ matrices with entire entries, satisfy $$A(z)A^*(z)+B(z)B^*(z)=C(z)C^*(z) \tag{1} $$ for all $z \in \mathbb {C}$ then $A=C(z)M$, $B=C(z)N$ for constant $n \times n$ matrices $N$ and $M$ with $MM^*+NN^*=I$.
Proof:
By the lemma there exists an entire matrix $C'(z)$ s.t $C'(z)C(z)=C(z)C'(z)=\det(C(z))$.
Multiply equation (1) on the left by $C'$ and the right by $C'^*$. This gives
$$a(z)a^*(z)+b(z)b^*(z)=|\det(C(z))|^2 I $$
where $a(z)=C'(z)A(z)$, $b(z)=C'(z)B(z)$, both matrices with entire entries.
If $a(z)=(a_{ij}(z))$, $b(z)=(b_{ij}(z))$ then the main diagonal of the LHS, $L(z)=(l_{ij})$ is given by $$l_{ii}=\sum_{j=1}^n(|a_{ij}|^2+|b_{ij}|^2)=|\det(C(z))|^2. \tag{2}$$
Hence for all the entire functions $a_{ij}$ and $b_{ij}$ we have $$|a_{ij}|\leq |\det(C(z))|.$$ and $$|b_{ij}|\leq |\det(C(z))|$$ where $\det(C(z))$ is also an entire function.
Again by Liouville's Theorem we must have $|a_{ij}|=\lambda_{ij} \det(C(z))$ and $|b_{ij}|=\mu_{ij} \det(C(z))$ for some $\lambda_{ij}, \mu_{ij} \in \mathbb {C}.$
This clearly implies that $a=\det(C(z)) M$ and $b=\det(C(z)) N$ for constant $n \times n$ matrices $M=(\lambda_{ij})$ and $N=(\mu_{ij})$.
Hence by the definitions of $a$ and $b$,
$C'(z)A(z)=\det(C(z)) M$ and $C'(z)B(z)=\det(C(z)) N$.
Note that if $\det(C(z))=0$ then by equation (2) $A(z)=0$, $B(z)=0$ and hence $C(z)=0$. Since then $A(z)=C(z)I$ and $B(z)=C(z)I$ we may take $M=N=I$.
We can therefore assume henceforth that $\det(C(z))\neq 0$.
Multiplying both equations on the left by $C(z)$ gives $\det(C(z))A(z)=\det(C(z)) C(z) M$ and $\det(C(z))B(z)=\det(C(z)) C(z) N.$
Hence we have, as $\det(C(z))\neq 0$,
$A(z)=C(z) M$ and $B(z)= C(z) N.$
Substituting back into equation (1) gives
$$A(z)A^*(z)+B(z)B^*(z)=C(z)(MM^*+NN^*)C^*(z)$$$$=C(z)C(z)^*$$
for all $z$, which implies $MM^*+NN^*=I$ since $C(z)$ is invertible. $\blacksquare$
