Here is a simple approach using the ping-pong lemma as OP suggests.

Consider the action of $F = \langle a,b\rangle$ on itself by left multiplication. Let $X_k \subseteq F$ be the set of elements which start with $b^ka$ or $b^ka^{-1}$, in their reduced word representation. As the ping-pong lemma requires, these are disjoint; and for $j \neq k$ and $n \neq 0$, we can see that $(b^k a b^{-k})^n X_j \subseteq X_k$ by the standard multiplication of reduced words. The claim follows.

Under the hood this is essentially the same proof as Benjamin Steinberg’s answer, but the ping-pong lemma helps organise the required analysis of reduced words.

Here is a direct approach using the ping-pong lemma as OP originally requests, which has been gestured at in several comments but not yet offered as an answer.

Consider the action of $F = \langle a,b\rangle$ on itself by left multiplication. Let $X_k \subseteq F$ be the set of elements which start with $b^ka$ or $b^ka^{-1}$, in their reduced word representation. As the ping-pong lemma requires, these are disjoint; and for $j \neq k$ and $n \neq 0$, we can see that $(b^k a b^{-k})^n X_j \subseteq X_k$ by the standard multiplication of reduced words. The claim follows.

Under the hood this is essentially the same proof as Benjamin Steinberg’s answer, but the ping-pong lemma helps organise the required analysis of reduced words.