I had a devil of a time figuring out the previous answer, but I now believe it to be fundamentally sound. Based on that answer, I am providing another one with a number of gaps filled in.

First note that we can assume WLOG that $a = 1$. Under this assumption we will show that either the real or the imaginary part of $\sum_{n=1}^\infty \frac {a_n} {n^s}$ diverges, where $s = 1 + it$ and $t \neq 0$. We also assume WLOG that $t > 0$. (Otherwise, replace $t$ by $-t$ in appropriate places below.) Put $a_n = u_n + iv_n$, where $u_n \rightarrow 1$ and $v_n \rightarrow 0$. Note that $$\Re(\frac {a_n} {n^s}) = (u_n \cos (t \log (n)) + v_n \sin (t \log (n))) / n$$ and $$\Im(\frac {a_n} {n^s})= (-u_n \sin (t \log (n)) + v_n \cos (t \log (n))) / n.$$

There must either be infinitely many values of n for which $|\cos (t \log (n))| > 1/2$ or infinitely many values of n for which $|\sin (t \log (n))| > 1/2$ . We assume the former. (Otherwise, switch the roles of sin and cos, and work with the imaginary part of the series instead of the real part; the necessary changes are fairly straightforward.) Likewise, WLOG we assume that there are inf. many values of n for which $\cos (t \log (n)) > 1/2$, as opposed to $< -1/2$. (Otherwise, replace cos by -cos and $\Re$ by $-\Re$ in appropriate places below.)

Now $\cos(x)$ is uniformly continuous on the real line. Select $\delta > 0$ such that if $|x - y| < \delta$, then $|\cos(x) - \cos(y)| < 1/4$ . We can also assume that $\delta < t$.

To prove divergence under the above assumptions, we will show that the sequence of partial sums of the real part of our series is not Cauchy. Let $M$ be a positive integer.

First find $M_1 > M$ such that for all $n > M_1$, $u_n > 1/2$ and $|v_n| < 1/16$. Choose $N$ an integer such that $N > \max(M_1, 2t / \delta)$ and $\cos (t \log (N)) > 1/2$ . Now $N \delta / 2t > 1$, whence $N \delta / t - N \delta / 2t > 1$. Thus, we can select $K$ an integer with $1 < N \delta / 2t < K < N \delta / t$ . Let $n$ be any integer with $n > N$. From calculus we know that
$$t(\log (n) - \log (N)) < t(n - N) / N.$$

Observe that if $N < n \leq N + K$ , then
$$t(\log (n) - \log (N)) < tK / N < \delta,$$
whence
$$|\cos(t \log (n)) - \cos(t \log (N))| < 1/4.$$
Therefore,
$$ \eqalign{&u_n \cos (t \log (n)) + v_n \sin (t \log (n))\geq u_n \cos (t \log (n)) - |v_n \sin (t \log (n))|\cr&\qquad\gt \cos (t \log (n)) / 2 - 1/16\cr&\qquad\geq (\cos (t \log (N)) - |\cos (t \log (n)) - \cos (t \log (N))|) / 2 - 1/16\cr&\qquad\gt (1/2 - 1/4) / 2 - 1/16 = 1/16.\cr}$$

It follows that
$$\eqalign{\sum_{n=N+1}^{N+K} \Re(\frac {a_n} {n^s}) &\gt \sum_{n=N+1}^{N+K} 1 / 16n \gt K / (16(N + K))= 1 / (16(N / K + 1))\cr& \gt 1 / (16(2t / \delta + 1))= \delta / (16(2t + \delta)) \gt \delta / 48t.\cr}$$

Thus, the sequence of partial sums of the real part of our series is not Cauchy.