Let $(a_n)_{n=0}^\infty \cdot (b_n)_{n=0}^\infty = (c_n)_{n=0}^\infty$ where $c_n = \sum_{k=0}^n a_k b_{n-k}$.
Following on from a previous question, I am considering sequences $(a_n)_{n=0}^\infty$ of complex numbers which satisfy
$$(a_n z^n)_{n=0}^\infty \cdot (a_n w^n)_{n=0}^\infty = (a_n (z+w)^n)_{n=0}^\infty \tag{1}$$
for all $z,w \in \mathbb{C}$. I know that one family of such sequences is $a_n = \frac{c^n}{n!}$ for some constant $c \in \mathbb{C}$, since
$$\sum_{k=0}^n (a_k z^k) (a_{n-k} w^{n-k}) = \sum_{k=0}^n \frac{c^k}{k!} z^k \frac{c^{n-k}}{(n-k)!} w^{n-k} = \frac{c^n}{n!} \sum_{k=0}^n \binom{n}{k} z^k w^{n-k} = \frac{c^n}{n!} (z+w)^n$$ as required. In fact, it is easy to see that the requirement
$$\sum_{k=0}^n a_k a_{n-k} z^k w^{n-k} = a_n (z+w)^n = \sum_{k=0}^n a_n \binom{n}{k} z^k w^{n-k}$$
is equivalent to
$$a_n \binom{n}{k} = a_k a_{n-k} \tag{2}$$
for all $n,k \in \mathbb{N}$ with $0 \leq k \leq n$.
Are the sequences given by $a_n = \frac{c^n}{n!}$ the only sequences satisfying (2)?
Yes, all such sequences are of the form $c^n/n!$, except for zero sequence $a_n=0$. Letting $k=1$ in $(1)$, you get $$ a_n=\frac1{n}a_1a_{n-1}, $$ for all $n\ge 1$. You can then prove by induction that this implies $$ a_n=\frac{a_1^{n}}{n!}\qquad \text{ for all }n\ge 0. $$ For the base case, use $a_n\binom{n}k=a_ka_{n-k}$ with $n=k=0$ to get $a_0=a_0^2$, so $a_0$ is $0$ or $1$. If $a_0=0$, you get the zero sequence, and if $a_0=1$, the base case $a_0=(a_1)^0/0!$ is proven, and the inductive step is easy.