Generalizing $\sum_{n=1}^\infty \frac{\Gamma(n)\Gamma(x)}{\Gamma(n+x)} = \frac1{x-1}$ to double sum Using the definition of the beta function, it's easy to show that

$$\begin{align*}
\sum_{n=1}^\infty \frac{\Gamma(n)\Gamma(x)}{\Gamma(n+x)} &= \sum_{n=1}^\infty \operatorname{B}(x,n) \\[1ex]
&= \sum_{n=1}^\infty \int_0^1 t^{x-1} (1-t)^{n-1} \, dt \\[1ex]
&= \int_0^1 t^{x-1} \sum_{n=1}^\infty (1-t)^{n-1} \, dt \\[1ex]
&= \int_0^1 t^{x-2} \, dt \\[1ex]
&= \frac1{x-1}
\end{align*}$$

I was wondering if we could generalize this to the case of a double sum,

$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{\Gamma(m) \Gamma(n) \Gamma(x)}{\Gamma(m+n+x)}$$

I've considered rewriting
$$\frac{\Gamma(m) \Gamma(n) \Gamma(x)}{\Gamma(m+n+x)} = \begin{cases} \operatorname{B}(m,x) \operatorname{B}(m+x,n) \\ \operatorname{B}(n,x) \operatorname{B}(n+x,m) \\ \operatorname{B}(m,n) \operatorname{B}(m+n,x) \end{cases}$$
The third form seems the most useful, and applying the method above leads me to
$$\begin{align*}
\sum_{m=1}^\infty \sum_{n=1}^\infty  \operatorname{B}(m,n) \operatorname{B}(m+n,x) &= \int_0^1 (1-t)^{x-1} \sum_{m=1}^\infty \sum_{n=1}^\infty \operatorname{B}(m,n) t^{m+n-1} \, dt
\end{align*}$$
$\operatorname{B}$ is symmetric, so
$$\begin{align*}
\sum_{m=1}^\infty \sum_{n=1}^\infty \operatorname{B}(m,n) t^{m+n-1} &= \sum_{m=1}^\infty \operatorname{B}(m,m) t^{2m-1} + 2 \sum_{m=2}^\infty \sum_{1 \le n < m} \operatorname{B}(m,n) t^{m+n-1} \\[1ex]
\end{align*}$$
and we can rewrite the first sum as
$$\begin{align*}
\sum_{m=1}^\infty \operatorname{B}(m,m) t^{2m-1} &= \sum_{m=1}^\infty \frac{\Gamma(m)^2}{\Gamma(2m)} t^{2m-1} \\[1ex]
&= \sum_{m=1}^\infty \frac{t^{2m-1}}{\binom{2m}m} \\[1ex]
&= \frac{4\arcsin\left(\frac t2\right)}{\sqrt{4-t^2}}
\end{align*}$$
The subsequent integral over $t\in[0,1]$ seems doable if $x$ is a positive integer; not so sure otherwise. I also don't know what else could be done with the other sum.
Mathematica gives a result in terms of hypergeometric functions but is unable to simplify any further.
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{\Gamma(m)\Gamma(n)\Gamma(x)}{\Gamma(m+n+x)} = \frac1{x^2} {}_3F_2\left(\left.\begin{array}{c|c}1,1,x\\1+x,1+x\end{array}\right\vert 1\right)$$
 A: You may find a simple solution in (Almost) Impossible Integrals, Sums, and Series, Sect. 6.60, page $533$, showing that
$$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{\Gamma(i) \Gamma(j) \Gamma(x)}{\Gamma(i+j+x)}=\frac{1}{2}\left(\psi^{(1)}\left(\frac{x}{2}\right)-\psi^{(1)}\left(\frac{x+1}{2}\right)\right),$$
where $\psi^{(1)}(x)$ is Trigamma function.
A: I have not been able to simplify the last result, but I did manage to figure out the path from double sum to hypergeometric function.
It turns out that either alternative to the "third option" mentioned in OP are more useful. We have for instance
$$\frac{\Gamma(m) \Gamma(n) \Gamma(x)}{\Gamma(m+n+x)} = \frac{\Gamma(m) \Gamma(x)}{\Gamma(m+x)} \cdot \frac{\Gamma(m+x) \Gamma(n)}{\Gamma(m+n+x)}$$
and we use the result from the first part of OP, rewritten in terms of $\Gamma$.
$$\sum_{n=1}^\infty \frac{\Gamma(n) \Gamma(x)}{\Gamma(n+x)} = \frac1{x-1} = \frac{\Gamma(x-1)}{\Gamma(x)} \implies \sum_{n=1}^\infty \frac{\Gamma(n) \Gamma(m+x)}{\Gamma(m+n+x)} = \frac{\Gamma(m+x-1)}{\Gamma(m+x)} \qquad (1)$$
Then to more closely align the result with the hypergeometric function, we rewrite the $\Gamma$-s as Pochhammer symbols, in particular
$$\begin{cases}(1)_n = \Gamma(n+1) = n! \\ (x)_n = \frac{\Gamma(n+x)}{\Gamma(x)} \\ (x+1)_n = \frac{\Gamma(n+x+1)}{\Gamma(x+1)}\end{cases} \qquad (2)$$
So we have
$$\begin{align*}
\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{\Gamma(m) \Gamma(n) \Gamma(x)}{\Gamma(m+n+x)} &= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{\Gamma(m) \Gamma(x)}{\Gamma(m+x)} \cdot \frac{\Gamma(m+x) \Gamma(n)}{\Gamma(m+n+x)} \\[1ex]
&= \sum_{m=1}^\infty \frac{\Gamma(m) \Gamma(x)}{\Gamma(m+x)} \cdot \frac{\Gamma(m+x-1)}{\Gamma(m+x)} & (1) \\[1ex]
&= \sum_{m=0}^\infty \frac{\Gamma(m+1) \Gamma(x)}{\Gamma(m+x+1)} \cdot \frac{\Gamma(m+x)}{\Gamma(m+x+1)} \\[1ex]
&= \sum_{m=0}^\infty \frac{\Gamma(m+1) \Gamma(x)^2}{\Gamma(m+x+1)} \cdot \frac{\Gamma(m+x)}{\Gamma(m+x+1) \Gamma(x)} \\[1ex]
&= \frac1{x^2} \sum_{m=0}^\infty \frac{\Gamma(m+1) \Gamma(x+1)^2}{\Gamma(m+x+1)} \cdot \frac{\Gamma(m+x)}{\Gamma(m+x+1) \Gamma(x)} & \Gamma(x+1)=x\Gamma(x) \\[1ex]
&= \frac1{x^2} \sum_{m=0}^\infty \frac{\Gamma(m+1) \cdot \frac{\Gamma(m+x)}{\Gamma(x)}}{\frac{\Gamma(m+x+1)^2}{\Gamma(x+1)^2}} \\[1ex]
&= \frac1{x^2} \sum_{m=0}^\infty \frac{[(1)_m]^2 (x)_m}{[(x+1)_m]^2} \cdot \frac1{m!} & (2) \\[1ex]
&= \frac1{x^2} \, {}_3F_2 \left(\left.\begin{array}{c}1,1,x\\x+1,x+1\end{array}\right\vert 1\right)
\end{align*}$$
