As Lee points out, your example seems to be a solution to the question as stated. Perhaps you want such $x$ to exist arbitrarily far from the identity? For instance, in a lamplighter, I would expect that for any sequence $x_n$, we should have $\limsup(d(x_n,J_S)/|x_n|)\leq 1/4$.

What does "projection" mean in a delta hyperbolic space?

I am not aware of any research into this interesting question. Perhaps the 1-dimensional version has been studied, since it is similar to the popular word game "Ghost". You might ask Chaim Goodman-Strauss.

Hey Andy. Suppose that $S$ and $T$ are generating sets. If two elements $g_1,g_2\in G$ differ by an element of $T$, then they can be connected by a path in the $S$-Cayley graph on which $\phi$ does not get much smaller than the minimum of $\phi(g_1)$ and $\phi(g_2)$. This means that, if $X_\phi$ is connected in the $T$-Cayley graph, then for $n$ sufficiently large, any two points of $\{g|\phi(g)>n\}$ are connected by a path in the $S$-Cayley graph on which $\phi$ stays non-negative. Any $g\in X_\phi$ can reach $\{g|\phi(g)>n\}$ in the $S$-Cayley graph since some $s\in S$ satisfies $\phi(s)>0$.

This is a great solution.

Why are the endpoints of the axis of an element of $\Gamma$ in $\mathbb{P}^1(K)$?

So I thought about this a bit and I didn't come up with an answer, but I would like to note that the following obvious strategy for producing a counterexample will never work. Given $\gamma\in PSL(2,\mathcal{O}_K)$, we may view gamma as a Moebius transformation of $\mathbb{C}\cup\infty$ in the usual way (i.e., $\frac{az+b}{cz+d}$). If $\gamma$ took the points $0,1,\infty$ to some points $p,q,r$ where $Im(p)$ and $Im(q)$ are positive but $Im(r)<0$, then $\mathcal{H}_2\cap\gamma\cdot\mathcal{H}_2$ would be two points. Unfortunately, this can never happen (!) for the fields $K$ in question.

@AndyPutman Why does $\gamma$ have to fix the intersection?