An example which is almost trivial: take the sequence $\{1,4,8,16,32\ldots\}$, i.e. powers of $2$ omitting two itself. For real part of $s>0$, the Dirichlet series is a convergent geometric series, summing as $$ \frac{1-2^s+4^s}{1-2^s} $$ which has a meromorphic continuation. The zeros are all purely imaginary, at $$ \frac{\pi i(6n\pm1)}{3\log(2)},\qquad \text{integer } n $$ if I've done the calculation correctly.
Building on this, one can sum $$ 1+2^{-s}+\sum_{k=3}^\infty 2^{-ks}=\frac{-2^{-2 s} \left(1-2^s+2^{3 s}\right)}{1-2^s}. $$ The numerator has a term which is a cubic in 2^s. A Mathematica computation then gives three infinite sets of zeros, one along the line with real part equal $$ \frac{\log \left(\sqrt[3]{\frac{2}{3 \left(9-\sqrt{69}\right)}}+\frac{\sqrt[3]{\frac{1}{2} \left(9-\sqrt{69}\right)}}{3^{2/3}}\right)}{\log (2)}\approx 0.405685 $$