Proof. This is essentially a triviality and a little bit of computation.

That $\mathcal{S}$ is an abelian group under $+$ is obvious; the only interesting point is noting that indeed $z$ is the identity of the group (the $0$ of the ring).

Many of the ring identities are also obvious. We will prove that $\varepsilon$ is the multiplicative identity, that $*$ is commutative and associative, that $*$ distributes over $+$, and that the units of the ring are as stated.

To see that $\varepsilon$ is the multiplicative identity, note that

and thus $\varepsilon*f=f$.

To see that $*$ is commutative, note that $f*g$ can also be written as

Commutativity is obvious from this representation of the operation.

Associativity follows similarly. Note that

If one expands $(f*(g*h))(n)$ similarly, the resulting sum is identical, so the two are equal.

Distributivity follows since

The units of the ring are simply the invertible functions; the entry on convolution inverses for arithmetic functions shows that the invertible functions are those functions $f$ with $f(1)\neq 0$.