Consider the following versions of Möbius inversion:


1. Let $(A,+)$ be an abelian group, and let $f$ and $g$ be functions $\mathbb N\rightarrow A$. Then $$\left((\forall n )\;g(n)=\sum\_{d|n}f(d)\right)\;\Longleftrightarrow\;\left((\forall n)\;f(n)=\sum\_{d|n}\mu(n/d)g(d)\right).$$
2. Let $R$ be a commutative ring with identity. Define $\mathcal A(R)=R^{\mathbb N}$ to be the set of all functions $\mathbb N\rightarrow R$. Thus $(\mathcal A(R),+,\ast)$ is a commutative ring with pointwise addition and Dirichlet convolution. For every $f\in\mathcal A(R)$ let $S\_f\in\mathcal A(R)$ be given by $S\_f(n)=\sum\_{d|n}f(d)$. Then $$(\forall f,g\in\mathcal A(R))\quad (g=S\_f\iff f=\mu\ast g).$$ Hence,
$$\left((\forall n )\;g(n)=\sum\_{d|n}f(d)\right)\;\Longleftrightarrow\;\left((\forall n)\;f(n)=\sum\_{d|n}\mu(n/d)g(d)\right).$$


The formulas themselves look the same; however, the second one naturally arises from the properties of the ring $\mathcal A(R)$, while the first does not come from such a richer structure -- it is simply proved directly from the definitions. I'm wondering if there is a common generalization of these formulas or some way of placing the first formula in a more general context like that of the second. 



Two versions of the Möbius inversion formula
One can work in the ring of operators that transform functions $\mathbb N \to A$ to functions $\mathbb N \to A$, for $A$ an abelian group - either a specific abelian group or operators defined on all ableian groups at once.


This is a noncommutative ring, but if you restrict to convolution operators, i.e. operators of the form $Tf(n) =\sum\_d f(d) g(n/d)$ for $g$ a $\mathbb Z$-valued function on $\mathbb N$ you get a commutative ring.


In this ring the elements corresponding to $g(n)=1$ and $g(n)=\mu(n)$ are inverses.


