The subgroup $\Gamma < \mathrm{SL}(2, \mathbb{R})$ generated by the matrices $ a = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} $ and $ b = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} $ is free of rank two. It acts on the upper half plane model of $\mathbb{H}^2$ via Mobius transformations. The orbit of $i$ gives the vertices of the Cayley graph; translates that differ by $a$, $b$, $a^{-1}$, or $b^{-1}$ are connected by a geodesic edge. It is an exercise to show that this gives the desired embedding. [Hint: build the Voronoi domain about $i$.]

The subgroup $\Gamma < \mathrm{SL}(2, \mathbb{R})$ generated by the matrices $ a = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} $ and $ b = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} $ is free of rank two. It acts on the upper half plane model of $\mathbb{H}^2$ via Mobius transformations. The orbit of $i$ gives the vertices of the Cayley graph; translates that differ by $a$, $b$, $a^{-1}$, or $b^{-1}$ are connected by a geodesic edge. It is an exercise to show that this gives the desired embedding. [Hint: build the Voronoi domain about $i$.]

A few remarks.

1. As YCor points out, no embedding can be isometric.
2. There are "quasi-isometric" embeddings of the Cayley graph into $\mathbb{H}^2$, but this one is not, as the generators are parabolic.
3. By taking finite index subgroups you can obtain equally nice actions of higher rank free groups on the hyperbolic plane.