Let me just replicate YCor's comment in the link that was given in the comments, with more details :
in a free product $C\ast D$ with $C,D$ nontrivial, the intersection of any two nontrivial normal subgroups is nontrivial.
Indeed, let $H,K$ be two nontrivial normal subgroups of $C*D$. I'll assume for simplicity that $|C|,|D|$ are large enough so that for each $x,y$ there is a nontrivial $z$ with $z^{-1}\neq x, z\neq y$. For instance $|C|, |D|\geq 4$ is good enough (take $x,y$, then there are at most two nontrivial $z$ such that $z=x$ or $z^{-1}=y$ : $x$ and $y^{-1}$; so if $|G|\geq 4$, any nontrivial element different from $x$ and $y^{-1}$ works)
Let me say that $c_1d_1\dots c_nd_n$ is the reduced form of an element of $C*D$ if $c_i\in C, d_j\in D$, and the only $c_i,d_j$ allowed to be $1$ are $c_1$ and $d_n$. Clearly, if $x=c_1d_1\dots c_nd_n$ is the reduced form of $x$, and $n\geq 2$, then $x\neq 1$ in $C*D$ ("clearly" here is to be understood as : it's a classical property of free products), moreover, if $n=1$ this is $1$ if and only if $c_1=d_1=1$.
Now let $x= c_1d_1\dots c_rd_r \in H, y=c'_1d'_1\dots c'_sd'_s \in K$ be nontrivial elements, with obvious notations, written in reduced form. The point will be that $[x,y]\in H\cap K$ (this is obvious by normality), and that, up to changing $x,y$ a bit, this can't be the trivial element.
Now up to conjugation by some element of $C$, one may assume $d_r = 1$ and $c_1\neq 1$ (this can be done by the hypothesis on $|C|$ - and $c_r \neq 1$, but that follows from the reduced form); and up to conjugation by some element of $D$, $c'_1 = 1$ , $d_s'\neq 1$ (using the cardinality hypothesis on $D$ - and $d_1' \neq 1$, but again this follows from the reduced form).
So with these hypotheses $$[x,y] = \color{red}{c_1d_1\dots c_r}\color{blue}{d_1'\dots c_s'd_s'}\color{red}{c_r^{-1}\dots d_1^{-1}c_1^{-1}}\color{blue}{d_s'^{-1}c_s'^{-1}\dots d_1'^{-1}},$$ which is written in reduced form, and is thus $\neq 1$. Therefore $[x,y]\in H\cap K\setminus\{1\}$.
Apply this to your supposed $A\times \{1\}, \{1\}\times B$ to get a contradiction.
I don't know if there's an easier argument, or a not-too-complicated argument that encapsulates the low cardinality cases, but I guess for these you have to go "by hand" somehow; or perhaps you can adapt this argument to these cases by working a bit more. In any case I didn't want to bother with these cases, and this argument works in most cases and is pretty painless so in any case it's interesting to share
Best Answer
No, because elements of $A$ and $B$ commute with each other. (Note I am identifying $A$ and $B$ with $A\times\{1_B\}$ and $\{1_A\}\times B$ respectively.) However in a free group the only elements which commute are powers of a common element. Any power of a mixed element $(a,b)$ where $a,b$ are nontrivial, stays mixed because $a$ and $b$ must have infinite order. Moreover any power of $(a,1_B)$ stays in $A$ and any power of $(1_A,b)$ stays in $B$. Thus elements of $A$ and $B$ cannot be powers of a common element.