It is a theorem of Schreier that any element of $G*H$ that has finite order must be conjugate to an element of $G$ or an element of $H$. (This is what Steve D notes in his comment).
To see this in the special case where $G$ is cyclic of order $n$ and $H$ cyclic of order $m$, let $G=\langle g\rangle$ and $H=\langle h\rangle$. Let $w$ be a reduced word in $G*H$ and assume that it is of finite order. Without loss of generality, let's say that it starts with $g$. Then we have $w = g^{a_1} h^{b_1} \cdots g^{a_k}h^{b_k}$ where $0\lt a_i\lt n$ for each $i$, and $0\lt b_i\lt m$ for $i=1,\dots,k-1$, with $0\leq b_i\lt m$.
If $b_k\gt 0$, then the length of $w^r$ is $2rk$, as there are no cancellations; so $w$ cannot be of finite order. Thus, $b_k=0$, and we have $w = g^{a_1}h^{b_1}\cdots h^{b_{k-1}}g^{a_k}.$ We can then conjugate $w$ appropriately so that either $a_i=0$ or $a_k=0$ (or both). Repeating the argument, we see that we must actually have $a_1+a_k = n$, so that $w$ is conjugate to $h^{b_1}g^{a_2}\cdots g^{a_{k-1}}h^{b_{k-1}}$. Now repeat the argument conjugating; eventually, we will end up with $g^{a_i}$ or $h^{b_j}$, so $w$ is conjugate to an element of $H$ or to an element of $G$.
So assume that G*H\cong G'*H'. Without loss of generality say $m\leq n$ and m',n\leq n'. If n\lt n' we have a contradiction, since G'*H' has an element of order n', but every element of finite order in $G*H$ has order dividing $n$ or $m$. So n=n', as desired.
Now we need to show that m=m'. We may assume m\leq m'.
The universal property of the coproduct tells us that the identity map $G\to G$ and the trivial map $H\to G$ induce a homomorphism $G*H\to G$ which restricts to the identity on $G$ and to the zero map on $H$. The kernel is then exactly the normal closure of $H$ of $G*H$, $K$; in particular, the normal closure of $H$ intersects $G$ trivially. Symmetrically, the normal closure of $G$ in $G*H$ intersects $H$ trivially.
The isomorphism G'*H'\to G*H maps the generator of H' to a conjugate of an element of $H$ or of $G$; and maps the generator of G' to a conjugate of an element of $H$ or of $G$. But they cannot both map to conjugates of elements of $H$, or both to conjugates of elements of $G$: because then the map $G*H\to G$ (respectively, the map $G*H\to H$) would be a nonzero map, but when composed with the isomorphism G'*H'\to G*H we would get the zero map, which is impossible. So if the generator of H' maps to a conjugate of an element (in fact, a generator) of $H$, then the generator of G' must map to a conjugate of an element of $G$, which proves that |G'|\leq |G|. This proves equality, since m\leq m'. If the generator of H' maps to a conjugate of an element of $G$ then n'\leq m\leq n=n', so we must have n=m=n'=m'.