Let $R = R^*$ be any graded ring which is graded-commutative in the sense of homological algebra, i.e. for homogeneous elements $x$ and $y$ we have $xy = (-1)^{|x| |y|} yx$. Consider the category of graded left $R$-modules. This has a tensor structure as follows. Any left $R$-module $M$ inherits a right action by $R$ via the formula $m\cdot r=(−1)^{∣m∣∣r∣}rm$. Using this, we can define a monoidal structure on left $R$-modules using the graded tensor product $M \otimes_R N$.
(Note that all of this really comes because graded abelian groups form a symmetric monoidal category under tensor product, using twist isomorphism $\tau(x \otimes y) = (-1)^{|x| |y|} y \otimes x$. In this category, $R$ is a commutative monoid object and the tensor is just defined by a standard coequalizer on modules.)
Now let $\mathcal{L} = R[1]$, by which I mean a shifted copy of $R$ so that the degree $n$ group $R[1]$ R[1]^n$is$R^{n+1}(grading cohomologically in order to align with the delicate sensibilities of the ag.algebraic-geometry tag). As a leftR$-module, it is free on a generator$e$with$|e|=-1$. Then this object is invertible, and tensor powers$\mathcal{L}^{\otimes n} = R[n]$are free on generators$e^{n}$for$n \in \mathbb{Z}$. At this point, one should verify for themselves that the ring $\Gamma_*(\mathcal L)$ is something like isomorphic to$R$as a graded ring. (Seriously, you should check this, especially if you usually take the attitude that "the signs just work themselves out". There may be a clever perspective that avoids sign issues, but the straightforward perspective is not so.) However, if you choose not to verify this: One then gets an identification $Hom_{gr. R-mod}(R, \mathcal{L}^{\otimes n}) \cong R[n]^{0} = R^n \cdot e_n$, and the multiplicative structure is given by $$r e^{|r|} \cdot s e^{|s|} = (-1)^{|r| |s|} (rs) e^{|rs|}.$$ In particular, this graded ring is noncommutative precisely when$R$is noncommutative, (which is most of the time). ADDED: You can verify that there is an isomporphism between$R$and this ring, given by the formula: $$\phi(r) = (-1)^{\binom{|r|}{2}} r \cdot e^r$$ There is no canonical sign switch if you use$\mathbb{Z}/2$-graded objects rather than$\mathbb{Z}$-graded objects (although mod-4 gradings are fine). 1 The following arose originally as a comment above and is being moved to an answer (per suggestion). Let $R = R^*$ be any graded ring which is graded-commutative in the sense of homological algebra, i.e. for homogeneous elements$x$and$y$we have$xy = (-1)^{|x| |y|} yx$. Consider the category of graded left$R$-modules. This has a tensor structure as follows. Any left$R$-module$M$inherits a right action by$R$via the formula$m\cdot r=(−1)^{∣m∣∣r∣}rm$. Using this, we can define a monoidal structure on left$R$-modules using the graded tensor product $M \otimes_R N$. (Note that all of this really comes because graded abelian groups form a symmetric monoidal category under tensor product, using twist isomorphism$\tau(x \otimes y) = (-1)^{|x| |y|} y \otimes x$. In this category,$R$is a commutative monoid object and the tensor is just defined by a standard coequalizer on modules.) Now let$\mathcal{L} = R[1]$, by which I mean a shifted copy of$R$so that the degree$n$group$R[1]$is$R^{n+1}(grading cohomologically in order to align with the delicate sensibilities of the ag.algebraic-geometry tag). As a leftR$-module, it is free on a generator$e$with$|e|=-1$. Then this object is invertible, and tensor powers$\mathcal{L}^{\otimes n} = R[n]$are free on generators$e^{n}$for$n \in \mathbb{Z}$. At this point, one should verify for themselves that the ring $\Gamma_*(\mathcal L)$ is something like$R$as a graded ring. (Seriously, you should check this, especially if you usually take the attitude that "the signs just work themselves out". There may be a clever perspective that avoids sign issues, but the straightforward perspective is not so.) However, if you choose not to verify this: One then gets an identification $Hom_{gr. R-mod}(R, \mathcal{L}^{\otimes n}) \cong R[n]^{0} = R^n \cdot e_n$, and the multiplicative structure is given by $$r e^{|r|} \cdot s e^{|s|} = (-1)^{|r| |s|} (rs) e^{|rs|}.$$ In particular, this graded ring is noncommutative precisely when$R\$ is noncommutative, (which is most of the time).