This is from p.$61$ in Wilf's "generatingfunctionology"
As a step to solving for the $b$'s in terms of the $a$'s
Given:
$a_n = \sum_{d\mid n}b_d$
Consider the Dirichlet power series generating functions of the two series $A(s)$ and $B(s)$, then
$A(s)=B(s)\zeta(s)$
This is based on a rule: $f(s)g(s)$ goes as Dirichlet power series generating function $\{\sum_{d\mid n} b_d c_{n/d}\}_{n=1}^\infty$
I would appreciate help with seeing the details of how this works Thanks
Lets look at how the multiplication rule works to answer your question. Let $A(s) = \sum a_n n^{-s}$ and let $B(s) = \sum b_n n^{-s}$. Then we have that,
\begin{align*} A(s)B(s) &= (a_11^{-s} + a_22^{-s} + a_33^{-s} + \cdots)(b_11^{-s} + b_22^{-s} + b_33^{-s} + \cdots)\\ &=a_1b_11^{-s} + (a_1b_2 + a_2b_1)2^{-s} + (a_1b_3 + a_3b_1)3^{-s}\\ &+ (a_1b_4 + a_2b_2 + a_4b_1)4^{-s} + \cdots \end{align*}
If you keep expanding and gathering terms you will see that the coefficient of $n^{-s}$ in $A(s)B(s)$ is $\sum_{d | n} a_{n/d} b_{d}$. If this isn't clear to you now, I encourage you to try calculating the coefficients of higher terms by hand.
Now returning to you original question, let $\zeta(s) = \sum_{n \geq 1} \frac{1}{n^s}$ this is the DGF of the sequence $c_n = 1$. Now we have that
$$\zeta(s)B(s) = \sum_{n\geq 1}\left(\sum_{d|n} c_{d/n}b_d\right)n^{-s} = \sum_{n\geq 1}\left(\sum_{d|n} b_d\right)n^{-s}$$
Since $c_n =1$.