In a recent Numberphile video on the Chebychev bias, Grant Sanderson (of 3blue1brown fame) uses what I guess would be the following theorem:
Given a multiplicative function $a$ where $\sum_n a(n)$ converges to a positive real number, consider the function $b$ defined as follows: $$ b(n) = \cases{\frac1k a(n) & if $n = p^k$ is a prime power \\ 0 & otherwise} $$ We have $$ \sum_{n = 1}^\infty b(n) = \ln\left(\sum_{n = 1}^\infty a(n)\right) $$
I have looked through my number theory books at home, and I haven't been able to find this result there. In the video they don't go into detail about it, they really just use it for some heuristics on the bias.
Is this the actual statement of the theorem? Or are there assumptions that I've missed (like convergence issues, or a need for complete multiplicativity)? And how do you prove it?
It is just taking the logarithm of an Euler product. Certainly there are convergence requirements, but that's irrelevant to understanding the structure of the formula itself. I'll explain the formula without dwelling on convergence issues. Everything to be done is valid on an open half-plane where the products converge absolutely, such as ${\rm Re}(s) > 1$ for Dirichlet $L$-functions.
The key background fact is that $$ \log\left(\frac{1}{1 - z}\right) = -\log(1-z) = \sum_{k \geq 1} \frac{z^k}{k} $$ when $|z| < 1$.
Now suppose $$ L(s) = \prod_p \frac{1}{1 - b_p/p^s}. $$ Then (ignoring convergence issues) $$ \log(L(s)) = \sum_p \log\left(\frac{1}{1 - b_p/p^s}\right) = \sum_p \sum_{k \geq 1} \frac{(b_p/p^s)^k}{k} = \sum_{p^k} \frac{b_p^k}{kp^{ks}}. $$ In the video, for prime $p$ we have $b_p = 1$ if $p \equiv 1 \bmod 4$, $b_p = -1$ if $p \equiv 3 \bmod 4$, and $b_2 = 0$.
This can be extended to Euler products with more than one linear factor in $1/p^s$ in the denominator. If $$ L(s) = \prod_p \frac{1}{(1 - b_{p,1}/p^s)\cdots (1-b_{p,d}/p^s)} $$ is an Euler product "of degree $d$" then (again ignoring convergence issues) $$ \log(L(s)) = \sum_p \sum_{i=1}^d\log\left(\frac{1}{1 - b_{p,i}/p^s}\right) = \sum_p \sum_{i=1}^d\sum_{k \geq 1} \frac{(b_{p,i}/p^s)^k}{k} = \sum_{p^k} \frac{b_{p,1}^k + \cdots + b_{p,d}^k}{kp^{ks}}. $$