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).

Here are some added reference and comments:

(a) Reference: this is the "famous" 1989 Pansu paper: Métriques de Carnot-Carathéodory et quasiisométries des espaces symétriques de rang un. (French) Ann. of Math. (2) 129 (1989), no. 1, 1–60.

(b) The continuum family consists of the 3-step-nilpotent simply connected Lie groups $G_t$ with the Lie algebra with basis $(e_i)_{1\le i\le 7}$ and brackets $12|4$ (meaning $[e_1,e_2]=e_4$), $13|5$, $23|6$, $16|7$, $25|t.7$, $34|(t-1).7$; it can be found in most lists. For, say $t$ transcendental, it's not QI to any discrete group. Magnin checked that $\frac{(t^2-t+1)^3}{t^2(t-1)^2}$ is an isomorphism invariant.

(c) Gromov proved that a discrete group with polynomial growth is virtually nilpotent. Lozert extended the result to locally compact groups, showing in particular that these are compact-by-Lie (hence compact-by-discrete in the totally disconnected case). Reference: V. Losert. On the structure of groups with polynomial growth. Math. Z. 195 (1987)
109-117. For groups acting on graphs (the totally disconnected case can boil down to it) Trofimov, reached the same conclusion: V. Trofimov. Groups with polynomial growth. Math USSR Sb 51 (1985) 405-418.

(d) It follows from Pansu's computation of $L^p$-cohomology and QI-invariance of the latter, that the groups $H_t=\mathbf{R}^2\rtimes\mathbf{R}$, for $t\ge 1$, with action $s\cdot (x,y)=(e^sx,e^{ts}y)$, are pairwise non-quasi-isometric. Reference: P. Pansu. Cohomologie Lp en degré 1 des espaces homogènes. Potential Anal. 27 (2007), no. 2, 151-165. Both steps for these groups were reproved differently in: Y. Cornulier, R. Tessera. Contracting automorphisms and Lp-cohomology in degree one. Ark. Mat. 49 (2011), no. 2, 295–324.

(Actually, it can be shown that for $t>1$, the Lie group $H_t$ is not QI to any vertex-transitive graph of finite valency. The main analytic ingredient is also Pansu 1989 (Cor. 6.9), but not the "famous" 1989 Pansu paper. Namely; Dimension conforme et sphère à l'infini des variétés à courbure négative, Ann. Acad. Sci. Fenn. Ser. A I Math. 14 (1989), no. 2, 177-212. See the discussion in my "quasi-survey" https://arxiv.org/abs/1212.2229, §19.6.3)

(e) For the Hilbert-Smith conjecture in dimension 2, it is asserted with a proof in Montgomery-Zippin's book but reading the proof I found a quite serious mistake. But everything could be fixed using results in Kolev, Boris Sous-groupes compacts d'homéomorphismes de la sphère. Enseign. Math. (2) 52 (2006), no. 3-4, 193-214. Also the 3-dim Hilbert conjecture by Pardon fixes this, although in a much more complicated way.