= Solution
Take both functions to be the constant arithmetic function $1$. Then
$$
(1\star1)(n)=\sum_{d\mid n}1=\tau(n),
$$
so the <divisor function> is multiplicative by part (a). For $s$ with $\operatorname{Re}s>1$, the relevant <Dirichlet series> are absolutely convergent, so their <Cauchy product> may be rearranged:
$$
\begin{aligned}
\zeta(s)^2
&=\sum_{a,b\geq1}\frac1{(ab)^s}
=\sum_{n\geq1}\frac1{n^s}\sum_{a\mid n}1
=\sum_{n\geq1}\frac{\tau(n)}{n^s}.
\end{aligned}
$$
Equivalently, Dirichlet convolution becomes multiplication of absolutely convergent Dirichlet series.
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