EDIT
The previous versions of this answer contained an error. I have now corrected the error and have verified its veracity.
Let's look at your last integral, which may be evaluated using a simple Bromwich contour and the residue theorem. Really, for $t \gt 0$,
$$\int_{c-i \infty}^{c+i \infty} du \, e^{u t} \Gamma(u)^2 = i 2 \pi \sum_{n=0}^{\infty} \operatorname*{Res}_{z=-n} e^{z t} \Gamma(z)^2 $$
Now we know that $\Gamma$ has the form, near $z=-n$ of
$$\Gamma(z) = \frac{a_{-1}}{z+n} + a_0 + O(z+n)$$
where $a_{-1} = (-1)^n/n!$ and
$$a_0 = \frac{(-1)^n}{n!} \left ( \sum_{k=1}^n \frac1{k} - \gamma \right ) = \frac{(-1)^n}{n!} \left ( H_n - \gamma \right ) $$
This result may be derived from the relation
$$\Gamma(-n+\epsilon) \Gamma(n+1-\epsilon) = (-1)^{n} \frac{\pi}{\sin{\pi \epsilon}} $$
which means that
$$\Gamma(-n+\epsilon) = \frac{(-1)^n}{n!} \frac1{\epsilon} + \frac{(-1)^n}{n!} \left ( H_n - \gamma \right ) + O(\epsilon)$$
Thus,
$$\operatorname*{Res}_{z=-n} \left [ e^{t z} \Gamma(z)^2 \right ] = a_{-1}^2 t + 2 a_{-1} a_0 = 2 \frac{H_n-\gamma+t/2}{n!^2}$$
Now, the integral we seek is
$$\int_{c-i \infty}^{c+i \infty} du \, e^{u t} \Gamma(u)^2 = i 2 \pi \sum_{n=0}^{\infty} \frac{2 (H_n-\gamma) + t}{n!^2} e^{-n t} $$
Now,
$$\sum_{n=0}^{\infty} \frac{z^n}{n!^2} = I_0 \left ( 2 \sqrt{z} \right ) $$
$$\sum_{n=0}^{\infty} H_n \frac{z^n}{n!^2} =K_0 \left ( 2 \sqrt{z} \right ) + \left ( \frac12 \log{z} + \gamma \right ) I_0 \left ( 2 \sqrt{z} \right )$$
where $I_0$ and $K_0$ are modified Bessel functions of the first and second kind, respectively, of zeroth order. These sums result from the respective expansions of $I_0$ and $K_0$ for small argument.
Subbing $z=e^{-t}$, we finally arrive at the value of the integral:
$$\int_{c-i \infty}^{c+i \infty} du \, e^{u t} \Gamma(u)^2 = i 2 \pi
\cdot 2 K_0 \left ( 2 e^{-t/2} \right ) $$
This is a two-sided transform, or a Mellin transform. (This was the original formulation of the problem.) The below derivation illustrates why.
ADDENDUM
I assumed that the first equation above is valid, but did not demonstrate how this is so. To see this, we must show that
$$R \int_{\pi/2}^{3 \pi/2} d\theta \, \left | \Gamma \left (R e^{i \theta} \right )^2 \right | e^{R t \cos{\theta}} $$
vanishes as $R \to \infty$.
Use Stirling's approximation:
$$\left |\Gamma \left (R e^{i \theta} \right )^2 \right | \sim \frac{2 \pi}{R} e^{2 R \log{R} \cos{\theta} -2 R \theta \sin{\theta}} e^{-2 R \cos{\theta}} $$
The term $R \log{R} \cos{\theta}$ dominates the other terms in the exponential, so that the behavior of this term is
$$\left |\Gamma \left (R e^{i \theta} \right )^2 \right | \sim \frac{2 \pi}{R} e^{2 R \log{R} \cos{\theta}} $$
Thus,
$$R \int_{\pi/2}^{3 \pi/2} d\theta \, \left | \Gamma \left (R e^{i \theta} \right ) \right |^2 e^{R t \cos{\theta}} \sim \frac{\pi^2}{R \log{R}} \quad (R \to \infty)$$
which indeed vanishes in the desired limit.
Note that the above criterion is independent of $t$. Thus, the transform is two-sided and not causal as was previously stated.
