= 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 gives
$$
\sum_{k\ge0}\left|\sum_{j=0}^ka^j\right|^2z^k
=\frac{1-z^2}{(1-z)^2(1-az)(1-a^{-1}z)}.
$$
For 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 proves
$$
\boxed{\sum_{n\ge1}\frac{|\sigma_{i\theta}(n)|^2}{n^s}
=\frac{\zeta(s)^2\zeta(s+i\theta)\zeta(s-i\theta)}{\zeta(2s)}.}
$$
Suppose now that $\zeta(1+i\theta)=0$. Necessarily $\theta\ne0$, 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 $\operatorname{Re}s>0$ the only possible remaining poles come from zeros of $\zeta(2s)$. Since the <Euler product> excludes zeros for $\operatorname{Re}(2s)>1$, these possible poles lie in $0<\operatorname{Re}s\le1/2$. In particular $f$ is regular for $\operatorname{Re}s>1/2$.
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 $f(\sigma)\ge1$ for real $\sigma>1/2$. But as $\sigma\downarrow1/2$, the numerator remains finite while $\zeta(2\sigma)\to+\infty$, forcing $f(\sigma)\to0$. This contradiction proves
$$
\boxed{\zeta(s)\text{ has no zero on }\operatorname{Re}s=1.}
$$
At 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.
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