= Solution
For $n\le x$, write $F(n)=F(x)-\int_n^xF'(t)dt$, multiply by $f(n)$ and sum. Interchanging a finite sum with the <integral> gives the <Abel summation formula>
$$
\sum_{n\le x}f(n)F(n)=S(x)F(x)-\int_1^xS(t)F'(t)\,dt.
$$
Apply it with $F(t)=t^{-s}$ and $S(t)=\sum_{n\le t}a_n$. If $\sigma>\alpha$, the boundary term is $O(x^{\alpha-\sigma})\to0$ and $\int_1^\infty|S(t)t^{-s-1}|dt$ converges. Therefore the <Dirichlet series> converges and
$$
A(s)=s\int_1^\infty S(t)t^{-s-1}dt=s\sum_{m\ge1}I_m(s).
$$
Choose $C$ with $|S(t)|\le Ct^\alpha$ for $t\ge1$. For $\sigma>\alpha+\delta$,
$$
|I_m(s)|\le C\int_m^{m+1}t^{\alpha-\sigma-1}dt\le Cm^{-1-\delta}.
$$
The <Weierstrass M-test> proves <uniform convergence> of $\sum I_m$ on this entire open half-plane. For fixed $s$, the same estimate with its actual $\sigma$ gives $I_m(s)=O(m^{\alpha-1-\sigma})$. Each $I_m$ is an <entire function>, since its <integral> has bounded positive $t$ range and can be differentiated in $s$ under an integrable bound on each <compact set>. Thus the locally uniform sum is <holomorphic>, and so is its product with $s$. This gives <holomorphy from polynomial growth of arithmetic partial sums> and proves holomorphy of $A$ for $\operatorname{Re}s>\alpha$. Multiplication by the unbounded factor $s$ does not assert <uniform convergence> of $A$ on a whole unbounded half-plane; local <uniform convergence> is what the holomorphy argument needs.
With $a_n=\mu(n)$ and the hypothetical bound, $A_\mu(s)$ is <holomorphic> for $\operatorname{Re}s>3/4$. On $\operatorname{Re}s>1$, the <Euler product> gives $\zeta(s)A_\mu(s)=1$. The <identity theorem> extends this as a meromorphic identity throughout the connected half-plane $\operatorname{Re}s>3/4$. Hence $\zeta$ has \b[no zeros in that half-plane]: at any zero, the product with the <holomorphic> $A_\mu$ would be zero. The <pole> at one is not a zero; $A_\mu$ has a compensating zero there.
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