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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 $R[1]$ 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 left $R$-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).
show/hide this revision's text 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 left $R$-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).