 Solution

For a finite set of primes $p\le y$, multiply the absolutely convergent local series. Unique prime factorization and multiplicativity, with $g(1)=1$, giveThe omitted integers all exceed $y$, so the absolute difference from $G(s)$ is at most ${\sum }_{n>y}|g(n)|n^{-\sigma }$, which tends to zero. Hence the finite products converge to $G(s)$:This proves the Euler product directly from absolute convergence, rather than treating an infinite formal multiplication as automatically valid.