Let $k$ be an algebraically closed field. If $S$ is a positively graded $k$-algebra which is finitely generated by $S_1$ over $S_0 = k$ then quasi-coherent sheaves on $\operatorname{Proj}S$ are equivalent to graded $S$-modules modulo the finite $k$-dimensional modules.
I have a positively graded finitely generated $k$-algebra whose generators are not necessarily in degree $1$. Does the statement above still hold? I've been googling around and every reference assumes that the ring is generated by $S_1$ but makes no mention of why. Is it just for convenience or is there a known counterexample?
If it fails I assume the functor that takes a graded module $M$ to it's sheaf $\widetilde M$ is at least still exact. Is it maybe still faithful? Or even full? Any references would be greatly appreciated.
$M\mapsto \widetilde{M}$ is exact: this is because the exactness is a local property and $\widetilde{M}(D_+(f))=M_{(f)}$ for any homogeneous $f$. Taking $M_{(f)}$ is easily seen to be exact.
It is faithfull modulo the part supported by the irrelevent maximal ideal $\mathfrak m$ exactly as in the case of $S$ generated by $S_1$. Indeed, if a graded $S$-linear map $\phi: M\to N$ is zero when passing to $\widetilde{M}\to \widetilde{N}$, then $\phi(M)_{(f)}=0$ for all homogeneous $f \in S_+$. So each element $x$ of $\phi(M)$ is annihilated by a power of $\mathfrak m$: if $\mathfrak m$ is generated by $f_1, \dots, f_n$ and $f_i^Nx=0$, then $\mathfrak m^{nN}x=0$.
It is "asymptotically" full for finitely generated $M$ as in the case of $S$ generated by $S_1$. What is important here is for any $d\ge 1$, $X:=\mathrm{Proj}(S)$ is also $\mathrm{Proj}(S(d))$ (non standard notation) where $S(d)=\oplus_{n\ge 0} S_{nd}$. As $S$ is finitely generated over $S_0$, $S(d)$ is finitely generated by $S(d)_1$ for some $d$. So using the special case where $S$ is generated by $S_1$, you see that $\oplus_{n\ge n_0} M_{dn}$ can be recovered from $\widetilde{M}$ when $n_0$ is big enough. As $M(d):=\oplus_{n\ge 0} M_{nd}$ and $\oplus_{n\ge n_0} M_{nd}$ differ by a finite-dimensional vector space, and $M/M(d)$ is supported in $\{ \mathfrak m\}$ (its $\widetilde{\kern 1pt}$ is trivial), you recover almost $M$ (not less than in the special case anyway).
I didn't check all the details, but I think you can work out them by yourself. Otherwise just tell.