[proofplan]
Write the two Dirichlet series as absolutely convergent ordinary series with terms $A_a=f(a)a^{-s}$ and $B_b=g(b)b^{-s}$. Absolute convergence implies that the double family $(A_aB_b)_{a,b\in\mathbb{N}}$ is absolutely summable, so its sum may be computed either as a product of sums or by grouping terms according to the product $ab$. The group with $ab=n$ is exactly indexed by divisors $d\mid n$, and its contribution is $(f*g)(n)n^{-s}$.
[/proofplan]
[step:Introduce the absolutely summable coefficient families]
For each $a\in\mathbb{N}$ define $A_a:=f(a)a^{-s}$, and for each $b\in\mathbb{N}$ define $B_b:=g(b)b^{-s}$. By hypothesis,
\begin{align*}
\sum_{a=1}^{\infty}|A_a|<\infty,
\qquad
\sum_{b=1}^{\infty}|B_b|<\infty.
\end{align*}
For finite sets $E,F\subset\mathbb{N}$, the product expansion gives
\begin{align*}
\sum_{a\in E}\sum_{b\in F}|A_aB_b|
=
\left(\sum_{a\in E}|A_a|\right)
\left(\sum_{b\in F}|B_b|\right)
\le
\left(\sum_{a=1}^{\infty}|A_a|\right)
\left(\sum_{b=1}^{\infty}|B_b|\right).
\end{align*}
Hence the double family $(A_aB_b)_{(a,b)\in\mathbb{N}^2}$ is absolutely summable.
[guided]
We first isolate the analytic issue from the arithmetic one. Define the scalar sequences
\begin{align*}
A_a:=f(a)a^{-s},
\qquad
B_b:=g(b)b^{-s}
\end{align*}
for $a,b\in\mathbb{N}$. The assumptions say exactly that
\begin{align*}
\sum_{a=1}^{\infty}|A_a|<\infty,
\qquad
\sum_{b=1}^{\infty}|B_b|<\infty.
\end{align*}
Now consider the two-indexed family $(A_aB_b)_{(a,b)\in\mathbb{N}^2}$. To justify any rearrangement or grouping later, we need absolute summability. For finite sets $E,F\subset\mathbb{N}$, finite distributivity gives
\begin{align*}
\sum_{a\in E}\sum_{b\in F}|A_aB_b|
=
\left(\sum_{a\in E}|A_a|\right)
\left(\sum_{b\in F}|B_b|\right).
\end{align*}
Since both partial sums are bounded by the corresponding full absolutely convergent series, we obtain
\begin{align*}
\sum_{a\in E}\sum_{b\in F}|A_aB_b|
\le
\left(\sum_{a=1}^{\infty}|A_a|\right)
\left(\sum_{b=1}^{\infty}|B_b|\right)
<\infty.
\end{align*}
Thus the double family is absolutely summable. This is the precise point where absolute convergence of both original Dirichlet series is used.
[/guided]
[/step]
[step:Compute the double sum as the product of the two Dirichlet series]
Let
\begin{align*}
S_f:=\sum_{a=1}^{\infty}A_a,
\qquad
S_g:=\sum_{b=1}^{\infty}B_b.
\end{align*}
For each $M,N\in\mathbb{N}$,
\begin{align*}
\left(\sum_{a=1}^{M}A_a\right)
\left(\sum_{b=1}^{N}B_b\right)
=
\sum_{a=1}^{M}\sum_{b=1}^{N}A_aB_b.
\end{align*}
Taking first $N\to\infty$ and then $M\to\infty$, using the absolute convergence of $\sum B_b$ and then of $\sum A_a$, gives
\begin{align*}
S_fS_g
=
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b.
\end{align*}
[guided]
For finite partial sums there is no convergence issue: ordinary distributivity gives
\begin{align*}
\left(\sum_{a=1}^{M}A_a\right)
\left(\sum_{b=1}^{N}B_b\right)
=
\sum_{a=1}^{M}\sum_{b=1}^{N}A_aB_b.
\end{align*}
Fix $M\in\mathbb{N}$. Since $\sum_{b=1}^{\infty}B_b$ converges absolutely, it converges, and therefore
\begin{align*}
\lim_{N\to\infty}\sum_{a=1}^{M}\sum_{b=1}^{N}A_aB_b
=
\sum_{a=1}^{M}A_a\sum_{b=1}^{\infty}B_b
=
\left(\sum_{a=1}^{M}A_a\right)S_g.
\end{align*}
Now let $M\to\infty$. Since $\sum_{a=1}^{\infty}A_a$ converges absolutely, it converges, so
\begin{align*}
\lim_{M\to\infty}\left(\sum_{a=1}^{M}A_a\right)S_g
=
S_fS_g.
\end{align*}
Thus the product of the two Dirichlet series is the double sum
\begin{align*}
S_fS_g
=
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b.
\end{align*}
The preceding step supplies absolute summability, which is what allows this double sum to be regrouped in the next step.
[/guided]
[/step]
[step:Group the double sum by the product $ab$]
For each $n\in\mathbb{N}$ define the finite divisor set
\begin{align*}
D_n:=\{d\in\mathbb{N}: d\mid n\}.
\end{align*}
The pairs $(a,b)\in\mathbb{N}^2$ satisfying $ab=n$ are exactly
\begin{align*}
(a,b)=\left(d,\frac{n}{d}\right)
\quad\text{with}\quad d\in D_n.
\end{align*}
Therefore
\begin{align*}
\sum_{\substack{a,b\in\mathbb{N}\\ ab=n}}A_aB_b
&=
\sum_{d\mid n}
f(d)d^{-s}g\!\left(\frac{n}{d}\right)\left(\frac{n}{d}\right)^{-s} \\
&=
n^{-s}\sum_{d\mid n}f(d)g\!\left(\frac{n}{d}\right) \\
&=
\frac{(f*g)(n)}{n^s}.
\end{align*}
Since the double family is absolutely summable, grouping by the disjoint finite sets $\{(a,b)\in\mathbb{N}^2:ab=n\}$ preserves the sum and gives
\begin{align*}
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}A_aB_b
=
\sum_{n=1}^{\infty}\frac{(f*g)(n)}{n^s}.
\end{align*}
The same absolute summability also gives
\begin{align*}
\sum_{n=1}^{\infty}\left|\frac{(f*g)(n)}{n^s}\right|
\le
\sum_{a=1}^{\infty}\sum_{b=1}^{\infty}|A_aB_b|
<\infty,
\end{align*}
so the Dirichlet series for $f*g$ converges absolutely at $s$. Combining this identity with the previous step proves the formula.
[/step]
