In the book "Groups, Graphs and Trees" by John Meier, Lemma 3.10 is the Ping Pong Lemma. He uses different assumptions than for example Wikipedia. Namely, he states that
Let $G$ be a group acting on a set $X$ and $S$ be a generating set for $G$. Let $\mathscr{S}=S\cup S^{-1}$. For every $s\in\mathscr{S}$, $X_s$ shall be a subset of $X$ and $p\in X\setminus\bigcup_{s\in\mathscr{S}}X_s$. Now if
- $s(p)\in X_s$ for every $s\in S$ and
- $s(X_t)\subset X_s$ (proper subset) for every $s\ne t^{-1}$,
then $G$ is a free group with basis $S$.
I can prove the result if the first condition actually holds for every $s\in \mathscr{S}$ (so for inverses too), because then if $w=w_1^{n_1}...w_{k-1}^{n_{k-1}}w_k^{n_k}$ is freely reduced ($w_i\ne w_{i+1}$, $w_i\ne w_{i+1}^{-1}$) we have $w(p)\in w_1^{n_1}...w_{k-1}^{n_{k-1}}(X_{w_k})\subset ... \subset X_1$, so $w\ne e$ because otherwise $p\in\bigcup_{s\in\mathscr{S}}X_s$. This also works if the last letter of $w$ is not an inverse.
I thought, this might be a misprint in the book, however I also couldn't come up with a counter example.
This was answered in the comments section. Briefly, either $w$ or $w^{-1}$ may be conjugated so that its final letter is in $S$, and then the argument applies.