 Solution

Use the convention $f(1)=1$ in the definition of a multiplicative arithmetic function. For a finite set of primes, multiplying the absolutely convergent local series gives the Dirichlet sum over integers supported on those primes, by unique prime factorization. If all primes up to $Y$ are included, any missing integer is larger than $Y$, so the difference from the full series is bounded by ${\sum }_{n>Y}|f(n)|n^{-\sigma }$, which tends to zero. This proves the Euler product in its absolute-convergence half-plane.

For coprime $m,n$, every divisor of $mn$ is uniquely a product of a divisor of $m$ and a divisor of $n$. Hence ${\sigma }_{\alpha }(mn)={\sigma }_{\alpha }(m){\sigma }_{\alpha }(n)$. The squared modulus is also multiplicative when $\alpha =i\theta$.

At a prime put $z=p^{-s}$ and $a=p^{i\theta }$, so $|a|=1$. Summing by the two divisor exponents givesFor example, expand the square as ${\sum }_{j,l\le k}{a}^{j-l}$ and sum $z^k$ from $k=\max (j,l)$ to obtain this rational expression. It remains valid for $a=1$. Absolute convergence for $\sigma >1$ follows from $|{\sigma }_{i\theta }(p^k)|\le k+1$ and convergence of ${\sum }_p{p}^{-\sigma }$. Multiplying the factors therefore provesSuppose now that $\zeta (1+i\theta )=0$. Necessarily $\theta \ne 0$, since $s=1$ is a pole. Complex conjugation gives a zero also at $1-i\theta$. In the numerator, the double pole at $s=1$ is canceled by those two zeros in the shifted factors. At $s=1\pm i\theta$, the possible simple pole of a shifted factor is canceled by the double zero of $\zeta (s{)}^2$. Thus on $\mathrm{Re}s>0$ the only possible remaining poles come from zeros of $\zeta (2s)$. Since the Euler product excludes zeros for $\mathrm{Re}(2s)>1$, these possible poles lie in $0<\mathrm{Re}s\le 1/2$. In particular $f$ is regular for $\mathrm{Re}s>1/2$.

The assumed nonnegative-series principle now makes its Dirichlet series converge and equal $f$ there. Its first coefficient is one and all coefficients are nonnegative, so $f(\sigma )\ge 1$ for real $\sigma >1/2$. But as $\sigma \downarrow 1/2$, the numerator remains finite while $\zeta (2\sigma )\to +\infty$, forcing $f(\sigma )\to 0$. This contradiction provesAt the real point $s=1$ the statement is understood as a pole rather than a finite nonzero value. The pole-location argument did not assume the zero-free boundary before proving it.