Let $M\succeq 0$ (i.e. $M_{ij}\geq 0$ for all $i,j$). No further conditions on $M$ such as irreducibility, aperiodicity, or what not. What is the formulation of the theorem in this case?
I believe that still the eigenvectors belonging to the largest eigenvalue are all non-negative. However the algebraic multiplicity of $\lambda_1$ is no longer 1, but rather I believe $n$. What is the statement here?
Edit:
Refined and more constrained question in a new post. It's okay for me if this one is put on hold.
This is discussed in some detail on the Wikipedia page. Each irreducible diagonal block of the $M$ satisfies the Perron-Frobenius theorem separately and the results of the theorem mostly carry over to the corresponding components of the eigenvectors. As you suspect, irreducible nonnegative matrices can have Perron-Frobenius roots with multiplicities other than one if they belong to different blocks. Within the same block, the magnitude can be the same but the phase can be different, and they are always roots of unity (see this section titled “Perron–Frobenius theorem for irreducible non-negative matrices”).