A complex arithmetic function is a map $f:\mathbb{Z}_{>0}\rightarrow\mathbb{C}$. Define $R$ as the set of all such functions. Let the sum of two such functions, denoted as $f_1+f_2$, be defined as
$$(f_1+f_2)(n)=f_1(n)+f_2(n)$$
and the convolution product, denoted as $f_1*f_2$, as
$$(f_1*f_2)(n)=\sum_{d\mid n}{f_1(d)f_2(\frac{n}{d})}$$
where the summation is taken over all positive divisors $d$ of $n$.
I have to prove that $R$, together with the summation and product as defined above, forms a (commutative) domain.
I guess that I have to prove all of the ring-axioms one by one (including commutative property) and then prove that $R$ has no zero-divisors. I think the zero would be the function $f_0(n)=0$ for all $n$ and with this, it is pretty straightforward to prove the abelian-group axioms for addition. The additive inverse of some $f(n)$ would be $(-f)(n)=-f(n)$ for all $n$. So far so good. Distributivity is also pretty easy to proof.
But now I have to prove the (other) multiplication axioms, which I simply can't seem to grasp. I'm not even sure what to choose as multiplicative identity, even though the straightforward guess would be to define it as the function $f_1(n)=1$ for all $n$ (please note that in the definition of the operants, the $f_1$ is an arbitrary function, not necessarily the unit). Can anyone help me understand what's going on here?
EDIT:
From a comment, I have gathered that $f_1(1)=1, f_1(n)=0$ for $n>1$ works as a $1$, proving the axiom of multiplicative identity.
I still need to prove the following:
Let $f,g,h\in R$, then
$$f*(g*h)=(f*g)*h,$$
$$f*g=g*f$$ and
$$f*g=f_0 \Rightarrow f=f_0\vee g=f_0$$