Is there any relationship between a Dirichlet series and the same series with the sequence "shifted" by one term? Suppose
$$ F(s) := \sum_{n=1}^{\infty} \frac{a_{n}}{n^{s}} $$
is a Dirichlet series for the sequence $a_{1}, a_{2}, \ldots\in\mathbb{C}$. Then let
$$ G(s) := \sum_{n=1}^{\infty} \frac{a_{n}}{(n+1)^{s}} $$
be the same series but with the sequence "shifted" by one index. Is there any known relationship between $F(s)$ and $G(s)$?
My question seems to be a subcase of this question, but I think my question should stand on its own, because the levels of generality here are different enough to warrant possibly different answers.
I'm okay with multiple suggestions.

There was an excellent suggestion made by user reuns that was later refined by Sangchul Lee.

Proposition. Let $\sigma_a$ be the abscissa of absolute convergence of $F$. Then for each $s$ with $\Re(s) > \sigma$,
$$ \sum_{k=0}^{\infty} \binom{-s}{k} [F(s+k) - a_1] = \sum_{n=2}^{\infty} \frac{a_n}{(n+1)^s}, $$
where the convergence of the series is absolute. Moreover,
$$ \lim_{\varepsilon \to 0^+} \sum_{k=0}^{\infty} \binom{-s}{k} e^{-\varepsilon k} F(s+k) = \sum_{n=1}^{\infty} \frac{a_n}{(n+1)^s} $$
(In other words, the proposed identity in OP holds in Abel summation sense.)

This is taken from this post here. For the sake of completeness, I will post their proof here as well.

To prove the first claim, note that
$$ \binom{-s}{k} = \frac{(-1)^k}{\Gamma(s)} \frac{\Gamma(s+k)}{\Gamma(k+1)} \sim \frac{(-1)^k}{\Gamma(s)} k^{s-1} \qquad \text{as } k \to \infty. $$
So, there exists a constant $C = C(s) \in (0, \infty)$ such that $\left|\binom{-s}{k}\right|\leq C (k + 1)^{\Re(s)-1}$ for all $k \geq 0$. Then by this bound and the Tonelli's theorem together, we get
\begin{align*}
\sum_{k=0}^{\infty} \sum_{n=2}^{\infty} \left| \binom{-s}{k} \frac{a_n}{n^{s+k}} \right|
&\leq C \sum_{k=0}^{\infty} \sum_{n=2}^{\infty} \frac{(k + 1)^{\Re(s)-1} |a_n|}{n^{\Re(s)+k}} \\
&\leq C \biggl( \sum_{k=0}^{\infty}  \frac{(k + 1)^{\Re(s)-1}}{2^k} \biggr)\biggl( \sum_{n=2}^{\infty} \frac{|a_n|}{n^{\Re(s)}} \biggr)
< \infty.
\end{align*}
So, by the Fubini's theorem,
\begin{align*}
\sum_{k=0}^{\infty} \binom{-s}{k} [F(s+k) - a_1]
&= \sum_{n=2}^{\infty} \sum_{k=0}^{\infty} \binom{-s}{k} \frac{a_n}{n^{s+k}} \\
&= \sum_{n=2}^{\infty} \frac{a_n}{n^s(1 + \frac{1}{n})^s}
= \sum_{n=2}^{\infty} \frac{a_n}{(n+1)^s}.
\end{align*}
The second claim then follows from the Abel's theorem for power series and
$$ \lim_{\varepsilon \to 0^+} \sum_{k=0}^{\infty} \binom{-s}{k} e^{-\varepsilon k} a_1
= \lim_{\varepsilon \to 0^+} \frac{a_1}{(1 + e^{-\varepsilon})^s}
= \frac{a_1}{2^s}. $$

