I think I can prove that, in fact, $M$ must be diffeomorphic to the trivial bundle over a compact homogeneous space, at least if $M$ is connected.

Let $G$ denote the isometry group.  Fix a point $p\in M$ and let $G_p$ denote the isotropy group.  Then $G_p$ is compact.  This follows because the action is proper (see, e.g, [this paper][1]), and then $G_p\times \{p\}\subseteq G\times M$ is the inverse image of the compact set $\{(p,p)\}\subseteq  M\times M$ under the map $G\times M\rightarrow M\times M$ given by $(g,m)\mapsto(gm,m)$.  Moreover, because the action is proper, we have a diffeomorphism $M\cong G/G_p$.

Writing $K$ for the maximal compact subgroup of $G$, we therefore have (up to conjugacy) inclusions $G_p\subseteq K\subseteq G$.  From this, one can form the homogeneous fibration $K/G_p\rightarrow G/G_p\rightarrow G/K$.

Now, $G/K$ is diffeomorphic to Euclidean space, which is contractible.  Thus, this fibration is trivial, so $M\cong G/G_p \cong K/G_p\times G/K$ as a manifold.


  [1]: https://arxiv.org/pdf/0811.0547.pdf