Divergence of Dirichlet series

The answer to your question is yes, the series must always diverge when $\Re (s)\in (0,1]$. The proof is a little delicate but here is the argument:

First of all to reiterate what you already mentioned, since the $a_n$'s have a non-zero limit, the abscissa of absolute convergence and the abscissa of convergence are both equal to $1$. Also the series clearly does not converge absolutely for $\Re (s)=1$, and it doesn't converge at all for $s=1$, so the only question is whether or not it converges conditionally for $s=1+it,~t\not= 0$.

Now for $t\not= 0$ let $$S_N=\sum_{n=1}^N\frac{a_n}{n^{1+it}}.$$ By a simple power series argument, for all $K,N\in\mathbb{N}$ with $K\le N/2$ we have that $$t(\log (N+K)-\log N)<\frac{ctK}{N},$$ for some universal constant $c>0$. Thus for $\epsilon >0$ if $N$ is large enough (depending on $t$) then choosing $$K=\left[\frac{\epsilon N}{ct}\right],$$ gives that $t(\log (N+K)-\log N)<\epsilon$.

Finally assume that $\epsilon$ is small and that $N$ is chosen as large as you need so that the quantities $a_n\exp(-it\log n)$ are all close to some constant $c'$, with $|c'|\approx |a|$, for all $N< n\le N+K.$ Then since the argument of the summand can not change by more than $\epsilon$ over this range we have that $$|S_{N+K}-S_N|=\left|\sum_{n=N+1}^{N+K}\frac{a_n\exp(-it\log n)}{n}\right|\gg |a|\sum_{n=1}^K\frac{1}{N+n}.$$ Pulling out the $N$ from the denominator gives $$|S_{N+K}-S_N|\gg \frac{|a|K}{N}\gg 1.$$ The final implied constant depends on $\epsilon$ and $t$, but both of these parameters are fixed (note that we don't have to let $\epsilon$ tend to $0$ to make this work, we just need it to be small). Therefore the sequence $S_N$ is not a Cauchy sequence and the Dirichlet series is not conditionally convergent for any value of $t$.

The underlying reason why the series don't converge is that the phase of the summand is not increasing fast enough to cause the necessary amount of cancellation.