Suppose by contradiction it is. Write it as $G/K$ where $G$ is the identity component of the isometry group and $K$ is compact, and $G$ acts faithfully on $G/K$. Since $G$ is connected, maximal compact subgroups are connected. Since $G/K$ is not contractible, $K$ is not maximal compact and it follows that maximal compact subgroups have codimension $1$. Since in noncompact simple Lie groups, maximal compact subgroups have codimension $\ge 2$, it follows that $G$ has no noncompact simple factor.
Suppose by contradiction that $G$ is not solvable. Then it has a simple compact connected subgroup $S$. Let $K'$ be a maximal compact subgroup. Then $K'\cap S$ has codimension $\le 1$ in $S$. Since a simple compact group has no subgroup of codimension $1$, it follows that $S\subset K'$. In turn, since $K$ has codimension $1$ in $K'$, by the same argument we deduce that $S\subset K$. Hence $S$ is contained in the intersection of all conjugates of $K$, which is trivial. Contradiction.
So $G$ is a connected solvable Lie group. Let $M$ be a closed connected normal subgroup of codimension $1$. If $MK=G$ then $M$ has smaller dimension and acts transitively, so we can argue by induction. Hence, assuming that $G$ has minimal dimension, we have $MK\neq G$, so $K\subset M$. Since in an abelian connected Lie group the intersection of all codimension 1 closed connected subgroups is trivial, we deduce that $K\subset \overline{[G,G]}$. The latter is nilpotent, and hence $K$ is a central torus in $\overline{[G,G]}$. Since the action of $G$ on the maximal torus of $\overline{[G,G]}$ is trivial, we deduce that $K$ is central, hence trivial. So $G$ is a 2-dimensional Lie group, and in particular is orientable.
(We have proved that if a noncompact connected surface can be endowed with a homogeneous Riemannian metric, then it is diffeomorphic to a Lie group, and hence to the plane, the cylinder. Of course various approaches to this result exist and some have already been mentioned in the comments.)
Best Answer
The only if direction fails. That is, there are $K$-equivariant vector bundles which are not homogeneous. For example, the Mobius band has the form $O(2)\times_{O(1)} \mathbb{R}$, and is not Riemannian homogeneous as you mention.
On the other hand, it seems the if direction is true. In fact, I think I can prove that $M$ must be diffeomorphic to the trivial bundle over a compact homogeneous space, at least if $M$ is connected.
Let $G$ denote the identity component of the isometry group (which still acts transitively). 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), 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 $H$ for the maximal compact subgroup of $G$, we therefore have (up to conjugacy) inclusions $G_p\subseteq H\subseteq G$. From this, one can form the fiber bundle $H/G_p\rightarrow G/G_p\rightarrow G/H$.
Now, $G/H$ is diffeomorphic to Euclidean space, which is contractible. Thus, this bundle is trivial, so $G/G_p$ is diffeomorphic to $(H/G_p)\times (G/H)$.