The answer is no. A quick answer can be done as follows:

(1) Pansu proved (1989) that two Carnot Lie groups are quasi-isometric if and only if they are isomorphic.

(2) There exists continuum many non-isomorphic 7-dimensional Carnot Lie groups.

If $Y$ is a proper, uniformly discrete and isometry-transitive metric space, then $Y$ is QI to its isometry group $G$, which is locally compact. Assuming it QI to a nilpotent Lie group in addition implies (by results of Gromov/Losert/Trofimov) that $G$, modulo a compact normal subgroup, is discrete and virtually nilpotent. There are only countably many QI classes. Hence at least one of the examples in (2) yields (for every choice of left-invariant Riemannian metric) a negative answer to your question.

There is an alternative using hyperbolicity rather polynomial growth, consisting of a continuum family of negatively curved homogeneous 3-folds. That they're not QI is also a result of Pansu. That a totally locally compact group QI to it has to be compact-by-discrete is an immediate consequence of the 2-dimensional Hilbert-Smith conjecture (which is an old theorem).