Let $g$ be a multiplicative function. Iwaniec and Fouvry claim the following identity on p. 273, identity (7.19). Why is this Euler product identity true?
$$-\sum_n \mu(n)g(n)\log n = \prod_{p} (1-g(p))(1-p^{-1})^{-1}$$
Even the case $g(n)=1/n$ is not clear to me.
$$F(s) = \prod_p \frac{1-g(p)p^{-s}}{1-p^{-1} p^{-s}} = \sum_n \frac{n^{-s}}{n} \sum_m \mu(m) g(m) m^{-s} = \sum_n n^{-s} \sum_{d | n} \mu(d) g(d) \frac{d}{n}$$ but we are told that for $s = 0$ it is also equal to : $$F(0) = -\sum_n \ln n \mu(n) g(n) $$
and in your link, $\mu(d) g(p)$ seems to be a particular case corresponding to $$\sum_{d | n} \mu(d) g(d) \frac{n}{d} = -\mu(n) \ln(n) g(n)$$
writing $G(s) = \sum \mu(n) g(n) n^{-s}$ we get that $G'(s) = -\sum \ln n \mu(n) g(n) n^{-s}$ so that if $\sum_{d | n} \mu(d) g(d) \frac{n}{d} = -\mu(n) \ln(n) g(n)$ it means $G(s) \zeta(s+1) = G'(s)$