Parabolic isometries on Gromov hyperbolic spaces

Gromov in his original 1987 book (Section 3.1) wrote a classification for arbitrary isometric group actions on hyperbolic spaces (with no further assumption) into 5 main classes. It goes at follows (the terminology is borrowed from here)

1: bounded: orbits are bounded

2: horocyclic: orbits are unbounded, $G$ acts with no hyperbolic isometry (hence there's a unique fixed boundary point)

3: lineal: there's a hyperbolic isometry and a fixed pair at infinity

4: focal: there's a hyperbolic isometry and a unique fixed point at infinity

5: general type: there's a hyperbolic isometry and no fixed point or pair at infinity.

According to Gromov, 1,2,3 are the elementary actions, i.e. the closure of an orbit in the boundary has at most 2 points, while in the non-elementary cases 4,5, this closure is uncountable.

If the action is cobounded, then case 2 (horocyclic action) can't occur. For the even more specific case of a discrete hyperbolic finitely generated group, case 4 (focal action) can't occur as well and Cases 1,3 are the virtually cyclic groups (finite and infinite).

On the other hand, all 5 cases occur for a group actions on trees, or on the hyperbolic plane (exercise).

Yes, there can be parabolic isometries in your sense. Examples can be obtained by "cone constructions", e.g. as follows:

Let $\mathbb{H}^2$ be the upper half-plane and let $X = (0,\infty) \times \mathbb{H}^2$ with the Riemannian metric $(dt)^2 + e^{-2t}\frac{\lvert dz\rvert^2}{y^2}$, which is a $\operatorname{CAT}(-1)$-space and hence geodesic and $\delta$-hyperbolic. Then $\operatorname{PSL}(2,\mathbb{R})$ acts freely by isometries on $X$ while it fixes a point in the boundary.

Any non-elementary Gromov-hyperbolic subgroup of $\operatorname{SL}(2,\mathbb{R})$ will yield an example.

A rather silly sufficient condition is to assume properness and cocompactness of the action, because the Svarc-Milnor lemma reduces this (up to quasi-isometric equivalence) to the case you already know.