Let $G$ be a group, $R$ a ring (not necessarily a field), and $M$ an $R$-module. Assume we have a group action $\rho:G \times M \to M$.
If there exists a $G$-invariant submodule $N \subseteq M$, is it always true that there exists a $G$-invariant complement $N'$ to $N$ such that we have a direct sum decomposition of $M$ as $$M=N \oplus N'$$
The proposition is when $R$ is a field. I can't figure out however where the assumption that $R$ is a field is necessary for the proof. Is the theorem true for general modules?
This isn't true at all. In fact, there need not exist any submodule ($G$-invariant or not) $N'\subset M$ such that $M=N\oplus N'$. Take, for instance, $R=M=\mathbb{Z}$ and $N=2\mathbb{Z}$. You can make this example equivariant for any group $G$ by just letting $G$ act trivially.
(By the way, it also isn't true in general when $R$ is a field. For instance, if $R$ is any field, $G=\mathbb{Z}$, and $G$ acts on $M=R^2$ by $1\cdot(u,v)=(u,u+v)$, then the subspace $N=\{(0,x):x\in R)\}$ is $G$-invariant but has no $G$-invariant complement. The case when it is true is if $G$ is a finite group and $R$ is a field whose characteristic does not divide $|G|$.)