Let $M$ be a graded module over an $\mathbb{N}$-graded ring $S$ and $S_+$ be the ideal of positive degree elements. Is it true that $M=0$ iff the homogeneous localization $M_{(\mathfrak p)}=0$ for all homogeneous prime ideals $\mathfrak p$ of $S$ not containing $S_+$?
The proof in the non graded case uses the fact that a nonzero element has a proper annihilator ideal. To follow the same proof I'd have to conclude that the annihilator does not contain all of $S_+$, which I'm not sure about, and then argue that the annihilator is contained in some prime graded ideal which does not contain all of $S_+$. We can assume $S$ is finitely generated by $S_1$ as an $S_0$-algebra if this helps.
EDIT: As shown by user26857 this is not true. I'd like to follow up by asking: if all homogeneous localizations at homogeneous primes $M_{(\mathfrak p)}=0$ is $M$ necessarily $0$?
No, this isn't true.
If $S=K[X]$ and $M=S/(X^2)$ then $M_{(0)}=0$ and $(0)$ is the only homogeneous prime of $S$ different from $S_+$. In general, I can't see any reason to exclude $S_+$ from such a question.
Edit. The answer to your edit is positive: take $x$ a homogeneous non-zero element of $M$ and note that $\operatorname{Ann}(x)$ is a homogeneous ideal, and therefore this is contained in a homogeneous prime ideal.