Let $s$ be a complex number with $Re(s)>1$. We know by the Euler product theorem that $\Pi_{\text{primes},p}(1-1/p^s)^{-1}=\zeta(s)$.
In this case, how do we ensure that $$\sum_{\text{primes},p}\log(1-1/p^s)^{-1}=\log \zeta(s),$$ that is what condition precisely allows us to move the infinite sum inside the logarithm to get the RHS which is the logarithm of the infinite product over primes?
Let $ s \in ( 1, \infty ) $ be real and denote by $ p _{ n} $ the $ n$th prime number. Then $$ \begin{aligned} \prod_{p}^{} \!\left( 1 - \frac{1}{ p ^{ s}} \right)^{-1} = \lim_{ n \to\infty} P _{ n} \end{aligned} $$ for $$ \begin{aligned} P _{ n} = \prod_{k = 1}^{n} \!\left( 1 - \frac{1}{ p _{ k} ^{ s}} \right)^{-1} .\end{aligned} $$ Now $$ \begin{aligned} \ln\!\left( \zeta ( s) \right) = \ln\!\left( \lim_{n \to \infty } P _{ n} \right) = \lim_{ n \to\infty} \ln\!\left( P _{ n} \right) = \lim_{ n \to\infty} \sum_{ k = 1}^{ n} \ln\!\left( ( 1 - 1 / p _{ k} ^{ s})^{-1} \right) \end{aligned} $$ by continuity of the natural logarithm. Notice that the lhs and rhs analytically extended to complex $s $ with $ \text{Re} \!\left( s\right) > 1$ so they must be equal on $ \text{Re} \!\left( s\right) > 1$.