Yes, the claim is true. It's a special case of a more general fact, that, in quite some generality, equivariant vector bundles are equivariantly locally trivial.

In your case, given $p$, there is a "slice" through $p$, a set $S\subset M$ containing $p$ such that $G\times S \to M$ is one-to-one and a homeomorphism onto an open subspace. By intersecting with a small enough nonequivariant neighborhood, we can assume that $L$ is trivial on $S$, and the action of $G$ then gives an equivariant trivialization of $L$ on $U = GS$. For a line bundle, an equivariant trivialization is equivalent to having a non-vanishing $G$-invariant section.

Slices exist pretty generally, as shown by Palais. See, for example, the writeup at the nLab.

Yes, the claim is true. It's a special case of a more general fact, that, in quite some generality, equivariant vector bundles are equivariantly locally trivial.

In your case, given $p$, there is a "slice" through $p$, a set $S\subset M$ containing $p$ such that $G\times S \to M$ is one-to-one and a homeomorphism onto an open subspace. By intersecting with a small enough nonequivariant neighborhood, we can assume that $L$ is trivial on $S$, and the action of $G$ then gives an equivariant trivialization of $L$ on $U = GS$. For a line bundle, an equivariant trivialization is equivalent to having a non-vanishing $G$-invariant section (assuming that the isotropy group, in this case the trivial group, acts trivially on the fiber at $p$).

Slices exist pretty generally, as shown by Palais. See, for example, the writeup at the nLab.