Dirichlet convolution preserves multiplicativity, and multiplicative inverses stay multiplicative

Since $f(1)=g(1)=1$, one has $(f\ast g)(1)=f(1)g(1)=1$. Now let $m,n$ be coprime positive integers. By unique prime factorization from The fundamental theorem of arithmetic: every integer $n\ge 1$ is a product of primes, and the factorisation is unique up to order — if ${\prod }_{i<r}{p}_i={\prod }_{j<s}{q}_j$ with every $p_i$ and $q_j$ prime, then $r=s$ and $q_i=p_{\pi (i)}$ for some $\pi \in \mathrm{Sym}(r)$, every positive divisor of $mn$ is uniquely of the form $ab$ with $a\mid m$ and $b\mid n$. In particular, every such pair $a,b$ is coprime.

By An arithmetic function has a Dirichlet inverse exactly when its value at $1$ is nonzero, the inverse $h$ exists because $f(1)=1\mathrm{\ne }0$, and $h(1)=1$.

Using the divisor splitting from step 1.1 and the multiplicativity of $f$ and $g$ on coprime pairs, $(f\ast g)(mn)={\sum }_{a\mid m,b\mid n}f(ab)g(mn\mathrm{/}ab)={\sum }_{a\mid m,b\mid n}f(a)f(b)g(m\mathrm{/}a)g(n\mathrm{/}b)=({\sum }_{a\mid m}f(a)g(m\mathrm{/}a))({\sum }_{b\mid n}f(b)g(n\mathrm{/}b))=(f\ast g)(m)(f\ast g)(n)$. Thus $f\ast g$ is multiplicative.

To prove $h$ is multiplicative, fix coprime positive integers $m,n$ and argue by strong induction on $mn$. If $m=1$ or $n=1$, then step 1.2 gives $h(mn)=h(m)h(n)$ because $h(1)=1$.

Now assume $m,n>1$ and that $h(uv)=h(u)h(v)$ for every coprime pair $u,v$ with $uv<mn$. Since $(f\ast h)(m)=\epsilon (m)=0$ and $(f\ast h)(n)=\epsilon (n)=0$, one has $0=({\sum }_{a\mid m}f(m\mathrm{/}a)h(a))({\sum }_{b\mid n}f(n\mathrm{/}b)h(b))={\sum }_{a\mid m,b\mid n}f(mn\mathrm{/}ab)h(a)h(b)$, where step 1.1 and multiplicativity of $f$ identify $f(m\mathrm{/}a)f(n\mathrm{/}b)$ with $f(mn\mathrm{/}ab)$. For each pair $(a,b)\mathrm{\ne }(m,n)$, step 1.1 gives $\gcd (a,b)=1$ and $ab<mn$, so the induction hypothesis yields $h(a)h(b)=h(ab)$. Reindexing the remaining pairs by $d=ab$ via step 1.1 therefore gives $0=h(m)h(n)+{\sum }_{\begin{array}{c}d\mid mn\\ d<mn\end{array}}f(mn\mathrm{/}d)h(d)$.

The inverse recursion from An arithmetic function has a Dirichlet inverse exactly when its value at $1$ is nonzero at $mn$ gives $0=h(mn)+{\sum }_{\begin{array}{c}d\mid mn\\ d<mn\end{array}}f(mn\mathrm{/}d)h(d)$. Comparing this with step 2.3 shows $h(mn)=h(m)h(n)$. Since the induction was on an arbitrary coprime pair, Strong (complete) induction yields multiplicativity of $h$.