A counterexample is provided by $c=1$ and $a_n=\sum_{d\mid n}(-1)^d$.
Indeed,
$$ D_{1,1,1}(s)=(2^{1-s}-1)\zeta(s)^2 $$
has a simple pole at $s=1$, but
$$ D_{1,1,2}(s)=\sum_{n=1}^\infty \frac{(-1)^na_n}{n^s}=(1-2^{1-s}+2^{1-2s})\zeta(s)^2 $$
has a double pole at $s=1$.

**EDIT:** Daniel asked in a comment how I came up with this sequence, so I try to explain. Vaguely, the original question asks the following: can we create a pole of a Dirichlet series by additive twists? In other words, can the coefficients of a Dirichlet series obey much less cancellation along arithmetic progressions than originally? The answer to this vague reformulation is obviously yes, e.g. there is a lot of cancellation in the formal sum $1+(-1)+1+(-1)+\dots$ but there is no cancellation in its formal subsums $1+1+1+\dots$ and $(-1)+(-1)+(-1)+\dots$. So a natural candidate for creating a pole by additive twists is $\sum_n (-1)^n/n^s$ which can be twisted to $\sum_n 1/n^s$. Indeed, here we create a pole at $s=1$, because the first series equals $(2^{1-s}-1)\zeta(s)$ but the second series equals $\zeta(s)$. Now this does not answer the original question since $(2^{1-s}-1)\zeta(s)$ has no pole at all. We can remedy this by considering $(1-2^{1-s})\zeta(s)^2$ instead, i.e. by convolving the sequence $(-1)^n$ with $1$. This is my final example above: I simply checked that it worked, i.e. it creates a double pole from a single one!