Local triviality of principal bundles

Suppose I define a principal $G$-bundle as a map $\pi: P \to M$ with a smooth right action of $G$ on $P$ that acts freely and transitively on the fibers of $\pi$. Does it follow that $P$ is locally isomorphic to $M \times G$ with the obvious right action of $G$ on $M \times G$? Let's suppose $M$ is a manifold.

I know that fiber bundles over a contractible set are trivial and a manifold is locally contractible, but I believe this statements refers to locally trivial fiber bundles and so will not apply to this case.

A related question is: if we have a fibration such that the base space is contractible and all fibers are homeomorphic, does it follow that the fibration is just the product of the base with the fiber?

Thanks!

For the first question - yes, at least if you suppose $P$ is a smooth manifold and, say, $G$ is a Lie group. For the principal $G$-bundles with your definition, just as with the regular one, being trivial is the same as admitting a section (in the section is $s$, map $s(x)$ to ($x$, unit of $G$), and use the action to define the rest of the trivializing map). Now to construct a section locally, near $x_0$ take any $s(x_0)$ in the fiber above $x_0$. Pick an auxiliary Riemann metric near take the orthogonal subspace to the tangent of the fiber at $s(x)$. Exponential map will give you a section locally.

In other categories you would need to construct the section $s$ in a different manner. I think this can be done in the category of topological manifolds. Not sure about more general cases, but it would seem ok. Maybe for CW complexes you can go cell by cell?