Convention: rings are unital and commutative.
Let $S=\bigoplus_{d\geq 0}S_d$ be a graded ring. For $n\in\mathbb Z$ let $S(n)$ be the graded $S$-module defined by $S(n)_d=S_{n+d}$. It is claimed in Stacks Project (2.27.2) that the canonical map
$$
\phi:S(n)\otimes_S S(m)\to S(n+m),x\otimes y\mapsto xy
$$
is an isomorphism. Surjectivity for me is clear(*). Injectivity not. Since the tensor product commutes with direct sums, we have
$$
S(n)\otimes_S S(m)=\bigoplus_{i,j}S_{n+i}\otimes_S S_{m+j}
$$
while
$$
S(n+m)=\bigoplus_d S_{n+m+d}.
$$
Let's assume $n=m$. If we take $i\neq j$, then we have the direct summand
$$
(S_{n+i}\otimes S_{n+j})\oplus (S_{n+j}\otimes S_{n+i})\subset S(n)\otimes_S S(m).
$$
Clearly then $(x\otimes y)\oplus (-y\otimes x)$ is mapped to zero. Where am I making a mistake? (assuming the claim that we have an isomorphism is correct)
(*) Thanks to thomasz pointing out that surjectivity isn't clear at all, here is a counter(**) example:
Take $S=k[x,y,z]$ with $\deg k=0,\deg x=1,\deg y=2,\deg z=3$. Then $S(1)\otimes_S S(2)\to S(3)$ doesn't reach $z\in S(3)$.
(**) This example doesn't work, as pointed out in the comments. I assumed we have an $\mathbb N$-grading, but in Stacks Project (1.10.56) we have the following convention: graded rings have an $\mathbb N$-grading, while graded modules have a $\mathbb Z$-grading. The twist operation is defined for graded modules and yields a graded module. Thus $S(n)$ is $\mathbb Z$-graded (meaning that also for $d<0$ we have $S(n)_d=S_{n+d} $). With this in mind, surjectivity is clear; we can reach $x\in S(n+m)_d$ by $x\otimes 1\in S(n)_{m+d}\otimes S(m)_{-m}$).
Best Answer
My counter-example for injectivity doesn't work: I forgot that we are tensoring over $S$ (and not for example over $S_0$), and therefore $$ x\otimes y+ (-y\otimes x)=x\otimes y- x\otimes y=0. $$ Also, I couldn't use that the tensor-product commutes with the direct sum, because $S_n$ is an $S_0$-module and not an $S$-module.
I see now that $\phi$ is in fact an isomorphism; its inverse is given by $$ \phi^{-1}:S(n+m)\to S(n)\otimes_S S(m),x\mapsto x\otimes 1. $$ Note that this is an $S$-module map. It is also a graded map, since it maps $x\in S(n+m)_d=S_{n+m+d}$ to $x\otimes 1\in S(n)_{m+d}\otimes S(m)_{-m}$ which has degree $m+d-m=d$.