The answer is no.
Since the Hilbert cube is compact and locally contractible, such a group would be a locally contractible locally compact group. And every locally contractible locally compact group is Lie (i.e., locally homeomorphic to $\mathbf{R}^d$ for some integer $d<\infty$).
For a reference
Szenthe, J.
On the topological characterization of transitive Lie group actions.
Acta Sci. Math. (Szeged) 36 (1974), 323–344. Link
Theorem 3 there: Let $G$ be a locally compact group and $H$ a closed subgroup such
that the coset space $G/H$ is locally contractible. Then $G/H$ is a free [=disjoint] union of manifolds
which are coset spaces of Lie groups.
(I've seen it attributed, when $H=1$ to earlier work of Gleason and Montgomery-Zippin without precise reference.)
Edit: Taras Banakh mentions in a comment that the Szenthe's proof has a gap, and that this gap is fixed independently in:
S. Antonyan, T. Dobrowolski, Locally contractible coset spaces.
Forum Math. 27 (2015), no. 4, 2157–2175. DOI link
K. Hofmann, L. Kramer. Transitive actions of locally compact groups on locally contractible spaces. J. Reine Angew. Math. 702 (2015), 227–243 (+ erratum 245–246). DOI link ArXiv link
Also, user Tyrone mentions that a negative solution to the OP's question (not addressing general locally contractible locally compact groups) is the statement of Theorem 3.1 in
A. Fathi, Y. Visetti. Deformation of open embeddings of Q-manifolds. Trans. Amer. Math. Soc. 224 (1976), no. 2, 427–435 (1977). link at AMS site DOI link