Let $X$ be the planar graph
 
 - whose vertices are the points $(i,j)$ for $i \in \mathbb{Z}$ and $j \in \{-1,0,1\}$;
 - whose edges connect $(i,0)$ with $(i+1,0)$ for every $i \in \mathbb{Z}$ and $(i,\pm 1)$ with $(i,0)$ for every $i \in \mathbb{Z}$.

In other words, $X$ is a bi-infinite line with top and bottom neighbours added to every vertex. 

Of course, $X$ is a locally finite hyperbolic graph (it's a tree). It is not difficult to verify that its isometry group is the unrestricted wreath product
$$\mathbb{Z}/2\mathbb{Z}\ \mathrm{wr} \ \mathbb{D}_\infty = \left( \prod\limits_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \rtimes \mathbb{Z} \right) \rtimes \mathbb{Z}/2\mathbb{Z}.$$
Each $\mathbb{Z}/2\mathbb{Z}$ from the product corresponds to an isometry that swaps the top and bottom neighbours $(i,1)$ and $(i,-1)$ for some $i \in \mathbb{Z}$ but fixes all the other vertices. The $\mathbb{Z}$ corresponds to the obvious translation. And the right $\mathbb{Z}/2\mathbb{Z}$ corresponds to a left-right reflection.

The boundary $\partial X$ of $X$ contains only two points, and the action of $\mathrm{Isom}(X)$ on $\partial X$ corresponds to the projection onto the right factor $\mathbb{Z}/2\mathbb{Z}$. Consequently, the kernel of the action is the unrestricted wreath product
$$\mathbb{Z}/2\mathbb{Z} \ \mathrm{wr} \ \mathbb{Z}:= \prod\limits_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z} \rtimes \mathbb{Z},$$
which is uncountably infinite.