We find an integral expression for the sum (the $u$ integral below) without appealing to the properties of special functions.
We have $\begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} &=& \frac{1}{\Gamma(s+1)} \underbrace{\left(1+\frac{1}{s+1}+\frac{1}{(s+1)(s+2)} + \ldots\right)}_{f(s)}. \end{eqnarray*}$ The series $f(s)$ is a simple example of an inverse factorial series. Such series were studied even in the 18th century by Nicole and Stirling and are dealt with, for example, in Whittaker and Watson's A Course of Modern Analysis.
One way to develop such a series is by successively integrating by parts the right hand side of $f(s) = \int_0^1 d\xi\, s(1-\xi)^{s-1} F(\xi),$ where $F(\xi)$ is some analytic function of $\xi$ and $\int_0^1$ is shorthand for $\lim_{\epsilon\to 0^+}\int_0^{1-\epsilon}$. One finds $\begin{eqnarray*} f(s) &=& F(0) + \frac{F'(0)}{s+1} + \frac{F''(0)}{(s+1)(s+2)} +\ldots. \end{eqnarray*}$ For details on the restrictions on $F(\xi)$, see Whittaker and Watson's 4th edition, $\S 7.82$.
For this problem we have $F^{(n)}(0) = 1$, so $F(\xi) = e^\xi$. Then $\begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} &=& \frac{f(s)}{\Gamma(s+1)} \\ &=& \frac{1}{\Gamma(s+1)} \int_0^1 d\xi\, s(1-\xi)^{s-1} e^\xi \\ &=& \frac{e}{\Gamma(s)} \int_0^1 du\, u^{s-1} e^{-u} \hspace{10ex}(\textrm{let }u=1-\xi) \\ &=& \frac{e}{\Gamma(s)} \gamma(s,1), \end{eqnarray*}$ where $\gamma(s,x)$ is the lower incomplete gamma function. Note that $\gamma(s,x) = \Gamma(s) - \Gamma(s,x)$, where $\Gamma(s,x)$ is the upper incomplete gamma function. Therefore, $\begin{eqnarray*} \sum_{n=1}^\infty \frac{1}{\Gamma(n+s)} &=& e\left(1-\frac{\Gamma(s,1)}{\Gamma(s)}\right), \end{eqnarray*}$ as claimed.
Thanks for the interesting question!