= Solution
For a finite set of primes $p\le y$, multiply the absolutely convergent local <series>. Unique <prime factorization> and multiplicativity, with $g(1)=1$, give
$$
\prod_{p\le y}\sum_{e\ge0}g(p^e)p^{-es}
=\sum_{\substack{n\ge1\\\text{every prime divisor of }n\le y}}g(n)n^{-s}.
$$
The 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 $G(s)$:
$$
\boxed{G(s)=\prod_pG_p(s)\qquad(\operatorname{Re}s>1).}
$$
This proves the <Euler product> directly from <absolute convergence>, rather than treating an infinite formal multiplication as automatically valid.
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