Disjoint-axis hyperbolic automorphisms satisfy ping-pong on a tree

Let $A_g$ and $A_h$ be the disjoint axes from [L1], and let $[p,q]$ be the unique geodesic joining them with $p\in A_g$ and $q\in A_h$. Removing $p$ from the tree leaves two components meeting the two rays of $A_g$, call them $X_g^{+}$ and $X_g^{-}$. Define $X_h^{+}$ and $X_h^{-}$ similarly at $q$. These four half-trees are pairwise disjoint.

Choose $m,n>0$ so large that $m\mathrm{\ell }(g)>d(p,q)$ and $n\mathrm{\ell }(h)>d(p,q)$. Then [L1] implies $g^m(T\setminus X_g^{-})\subseteq X_g^{+}$ and $g^{-m}(T\setminus X_g^{+})\subseteq X_g^{-}$, and similarly $h^n(T\setminus X_h^{-})\subseteq X_h^{+}$ and $h^{-n}(T\setminus X_h^{+})\subseteq X_h^{-}$. Indeed, a sufficiently long translation along an axis carries everything on one side of the basepoint across the bridge segment $[p,q]$ into the corresponding attracting half-tree.

Pick a vertex $x$ outside the four half-trees. Any reduced word in $g^{\pm m}$ and $h^{\pm n}$ sends $x$ into the domain attached to its first letter by step 2.1, and successive letters cannot bring it back because the four domains are disjoint. So no nonempty reduced word acts trivially. Therefore $g^m$ and $h^n$ freely generate a free subgroup of rank $2$.