Here is one way to prove it:
1) If the triangle group is uniform, then Milnor-Schwarz shows that it is word-hyperbolic (since it acts properly discontinuously and cocompacty by isometries on the hyperbolic plane).
2) If the triangle group is nonuniform, then we can argue as follows:
a) There is a single quasi-isometry class of nonuniform lattices in $SL_2(\mathbb R)$ (see, eg., Farb: "The quasi-isometry classification of lattices in semisimple Lie groups").
b) $SL_2(\mathbb Z)$ is word-hyperbolic (e.g. as being isomorphic to $\mathbb Z/4\mathbb Z \ast_{\mathbb Z/2\mathbb Z} \mathbb Z/6\mathbb Z$).
c) Therefore the index 2 subgroup in the triangle group consisting of the orientation-preserving isometries is word-hyperbolic.
d) Thus, the triangle group itself is word-hyperbolic.
However, the part 2) feels a bit overkill.
The infinite dihedral (diagram $\circ\stackrel{\infty}{—}\circ$) is the only one.
More precisely: if a Coxeter group $(W,S)$ is both virtually abelian and infinite hyperbolic then it's given by the above Coxeter system, possibly up to disjoint union with a Coxeter system of finite type.
Indeed, being hyperbolic and virtually abelian, it's virtually cyclic. It's known (Wall 1967) that a infinite virtually cyclic group is always finite-by-$\mathbf{Z}$ or finite-by-$D_\infty$. For a group generated by torsion elements, the first case is excluded.
For a group that is finite-by-$D_\infty$, there are two kinds of elements of order $2$: those in the finite kernel (which is the unique maximal finite normal subgroup), and those mapping to an element of order $2$ in the $D_\infty$ quotient. Call them of type (a) and (b) respectively.
Claim: every Coxeter generator of type (a) commutes with every Coxeter generator of type (b).
Granting the claim: those Coxeter generators of type (a) form a union of components in the Coxeter graph, and hence these form a finite direct factor. Hence we can reduce to the case when all Coxeter generators are of type (b). In $D_\infty$, any two distinct elements of order $2$ have product of infinite order. At least two Coxeter generators, say $u,v$ have distinct images in $D_\infty$. Suppose by contradiction that there is a third one $w$. If its image in $D_\infty$ is not the same as that of $u$ or $v$, then all $uv$, $uw$, $vw$ have infinite order. So $u,v,w$ form a triangle with $\infty$ edges, and this forms a group with a non-abelian free subgroup of index $2$. Otherwise, say $w$ has the same image as $u$. Then $u,v,w$ forms a triangle with two $\infty$ edges, and the corresponding subgroup is a free product $D_n\ast C_2$ with $D_n$ dihedral of order $\ge 4$, so again this has a non-abelian free subgroup of finite index.
For the claim: consider $u$ of type (b), not commuting with $w$ of type (a). Again, there exists $v$ of type (b), with an $\infty$ edge between $u$ and $v$. So we have a triangle with labels $\infty$, $n\ge 3$, and $m\ge 2$. This Coxeter group is an amalgamated product $D_{2n}\ast_{C_2} D_{2m}$ over a common generator. Since the amalgamated subgroup has index $\ge 2$ in one and index $\ge 3$ in the other, it also has a non-abelian free subgroup of finite index.
Best Answer
It does not follow that for groups which are not finitely generated the notion of hyperbolicity makes sense.
What follows, instead, is that for any pair $(G,S)$ such that $G$ is a group and $S$ is a generating set, the notion of hyperbolicity is well-defined: the Cayley graph $\Gamma(G,S)$ is certainly defined (and it is connected), and hyperbolicity of $\Gamma(G,S)$ certainly makes sense.
The trouble is that without any control on the cardinality of $S$, it is possible to have a group $G$ and two generating sets $S_1,S_2$ such that $\Gamma(G,S_1)$ is hyperbolic whereas $\Gamma(G,S_2)$ is not.
In fact, here's an example: $G = \mathbb Z^2$ with $S_1 = \{(\pm 1,0),(0,\pm 1)\}$ and with $S_2 = \mathbb Z^2$. The Cayley graph $\Gamma(G,S_1)$ is not hyperbolic whereas $\Gamma(G,S_2)$ is bounded and is therefore hyperbolic. That's an example where $G$ is in fact finitely generated, but one can easily cook up examples where $G$ is not finitely generated, e.g. $G = \mathbb Z^\infty$ with the "standard" generating set $S_1 = \{\pm e_i \mid i \ge 1\}$ is not hyperbolic, but with the generating set $S_2=G$ it is hyperbolic.
What happens in the finitely generated case is that if $S_1,S_2$ are two finite generating sets for the same group $G$, then $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are quasi-isometric to each other, and hyperbolicity is a quasi-isometry invariant, so $\Gamma(G,S_1)$ is hyperbolic if and only if $\Gamma(G,S_2)$ is hypebolic. Thus one says that a finitely generated group $G$ is hyperbolic if and only if some (hence any) finite generating set $S$ results in a hyperbolic Cayley graph $(G,S)$.
The summary statement is that hyperbolicity of a finitely generated group is well-defined independent of the choice of finite generating set, but take away those bold faced words and the statement becomes false.
ADDED: In answer to a followup question in the comments, every Gromov hyperbolic space (i.e. every geodesic metric space satisfying the thin triangle property) has a Gromov boundary.
And, given a group $G$ and two generating sets $S_1$, $S_2$, if $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ are both hyperbolic, then both of them do have a Gromov boundary. And if $S_1,S_2$ are both finite then one can use the quasi-isometry between $\Gamma(G,S_1)$ and $\Gamma(G,S_2)$ to deduce that their Gromov boundaries are homeomorphic; so in the finitely generated case you are free to speak about THE (homeomorphism class of the) Gromov boundary.
However, without the assumption that $S_1,S_2$ are both finite, it need not be true that the Gromov of $\Gamma(G,S_1)$ is homeomorphic to the Gromov boundary of $\Gamma(G,S_2)$.
For a counterexample, take your favorite infinite, finitely generated, hyperbolic group $G$. With any finite generating set $S_1$, the Gromov boundary is nonempty. With the infinite generating set $S_2=G$, the Gromov boundary is empty.
Anyway, it follows that to speak of THE Gromov boundary of $G$ is nonsensical in the infinitely generated case.