Here is how I would prove it using ping-pong. The relevant form of ping-pong is the following proposition:

Proposition. Let $X$ be a set and $G_1, G_2$ be groups acting on $X$ such that there exist two subsets $X_1, X_2\subset X$ with disjoint complements and nonempty intersection, such that
$$
\forall g_i\in G_i -\{1\}, g_i(X_i)\cap X_i=\emptyset.
$$
Then the group $\langle G_1, G_2\rangle$ of bijections of $X$ generated by $G_1, G_2$ is naturally isomorphic to $G_1\star G_2$.

Now, back to our hyperbolic group $G$. I will assume that $G$ is a non elementary hyperbolic group whose finite radical $W(G)$ is trivial. (Otherwise, see Yves' answer.) Given $g\in G$, let $Fix(g)$ denote the fixed-point set of $g$ in $\partial G$ (the Gromov boundary of $G$).

The following lemma answers Yves' question:

Lemma 1. If $g\in G-\{1\}$ has finite order then $Fix(g)$ has empty interior.

Proof. Suppose not. Then, since $W(G)=1$, $Fix(g)\ne \partial G$.

Recall:

Theorem. The set of pairs $(p_+, p_-)$ of attractive/repulsive fixed points of infinite order elements of $G$ is dense in $\partial G\times \partial G$.

Hence, one can find such fixed pair for an element $h\in G$ such that $p_+\notin Fix(g), p_-\in Fix(g)$. Then the sequence of conjugates
$$
g_n=h^n g h^{-n}, n\in {\mathbb N},
$$
contains infinitely many distinct elements (since the sequence of complements to $Fix(g_n)$ in $\partial G$ converges to the singleton $p_+$). At the same time, the sequence $(h^{-n})_{n\in {\mathbb N}}$ is uniformly close to the quasiconvex hull of $Fix(g)$. Since $g$ acts on this quasiconvex hull with bounded displacement, we obtain that the sequence of word norms $|g_n|$ is bounded. A contradiction. qed

Lemma 2. Let $g\in G$ be nontrivial, generating a finite cyclic group $C< G$. Let $h\in G$ be an infinite order element such that
$$
g^k(Fix(h))\cap Fix(h)=\emptyset
$$

for all $g^k\in C-\{1\}$. Then there exists $n$ such that the subgroup $\langle g, h^n\rangle$ of $G$ generated by $g$ and $h^n$ is naturally isomorphic to $C \star \langle h^n \rangle$.

Proof. Let $U$ be a small neighborhood of $Fix(h)$ in $\partial G$ such that
$$
g^k(U)\cap U=\emptyset
$$

for all $g^k\in C-\{1\}$ (such a neighborhood existents since $C$ is finite). Let $V= \partial G -U$. Then there exists $n\in {\mathbb N}$ such that for all nontrivial elements $f\in \langle h^n\rangle$ we have
$$
f(V)\cap V=\emptyset.
$$
It now follows from Tits' ping-pong that the subgroup $\langle g, h^n\rangle$ is naturally isomorphic to $C \star \langle h^n\rangle $. qed

Lemma 3. Given a nontrivial finite order element $g\in G$, there exists an infinite order element $h\in G$ such that $$
g^k(Fix(h))\cap Fix(h)=\emptyset
$$

for all $g^k\in \langle g \rangle -\{1\}$.

Proof. Let $U$ be a nonempty open subset of $\partial G -U$. Since $C=\langle g\rangle $ is finite, we can choose $U$ such that
for all nontrivial elements $f\in C$ we have
$$
f(U)\cap U=\emptyset.
$$
(For instance, one can use Lemma 1 here.)

On the other hand, by minimality of the action of $G$ on $\partial G$,
there exists an infinite order element $h\in G$ such that $Fix(h)\subset U$. (This also follows from the Theorem stated above.) qed

Now, Lemmas 2 and 3 imply:

Corollary. Let $g\in G$ be a nontrivial finite order element.
Then there exists an infinite order element $h\in G$ such that the subgroup $\langle g, h\rangle$ of $G$ is naturally isomorphic to $\langle g\rangle \star \langle h\rangle$. In particular, the product $gh$ also has infinite order.