Counterexample for Dirichlet product of two completely multiplicative functions. The below text is the proof of why Dirichlet product of two multiplicative functions are multiplicative:

  

It's obvious how the assumption of $(m,n)=1$ allowed the proof to be proceeded. However I am trying to find a counterexample to show that the Dirichlet product of two completely multiplicative functions is not be always completely multiplicative, but I couldn't succeed. Considering f and g to be power functions, $n^a$ and $n^b$ respectively, don't help since I can't find a way to decompose the double sum but it doesn't mean that $h(mn) \ne h(n)h(m)$. Is there any elementary counterexample?
If $p$ is a prime and $h = f * g$, $h(p) = f(1) g(p) + f(p) g(1) = g(p) + f(p)$ while $h(p^2) = f(1) g(p^2) + f(p) g(p) + f(p^2) g(1) = g(p)^2 + f(p) g(p) + f(p)^2$, so for $f * g$  to be completely multiplicative we'd need $f(p) g(p) = 2 f(p) g(p)$, and thus $f(p) = 0$ or $g(p) = 0$.
A counterexample is $I: \mathbb N\mapsto \mathbb N$ you have :
$$\forall n\in \mathbb N \quad \quad (I*I)(n)=nd(n) $$
and we know that $n\mapsto nd(n)$ is not completely multiplicative because $d$ the divisor counting function is not completely multiplicative.