Center of a free product

Well, you would have a hard time proving that a reduced word only commutes with the identity, because every word commutes with every power of itself, and of course, any element $a\in A$ will commute with all elements in $C_A(a)$, and any $b\in B$ with all elements in $C_B(b)$. So no element of $A*B$ has trivial centralizer! But you don't need a nontrivial element that doesn't commute with anything for the center to be trivial (good thing, since there is no such thing in any nontrivial group); you just need every nontrivial element to not commute with something.

You don't really need to consider a lot of cases: the only two things you have to worry a bit about are what to do if both of your groups have only two elements, and what the reduced word starts and ends with. For example, if $\mathbf{w}$ is a reduced word that starts and ends with an element of $A$, then just take $b\in B$, $b\neq e$ to get something it does not commute with, so $\mathbf{w}$ cannot be central. If it starts with an element of $A$ and ends with an element of $B$, then take a word $\mathbf{w}'$ that starts with an element of $A$ and ends with an element of $B$ (so that $\mathbf{w}\mathbf{w}'$ and $\mathbf{w}'\mathbf{w}$ are already reduced), and make $\mathbf{w}'$ either begin with something different from what $\mathbf{w}$ begins with, or end with something different from what $\mathbf{w}$ ends with; etc. You should have just 4 cases, and only two that can give you trouble if your groups are too small, in which case you can just stare at them directly.

Your argument for why $A*B$ is infinite is fine. You can even produce, explicitly, words of arbitrary length: just take $a\in A$, $b\in B$ with $a\neq e\neq b$, and consider $(ab)^n$, $n\in\mathbb{Z}$.