This is merely a reformulation of Abelmalek Abdessalam's answer, in a somewhat different language and with different references. It should be a comment to that answer, but it is unfortunately too long. Long story short: see Karlin, Total positivity, formula (2.10) in Section 1.2, with $u(t) = t$ and $\sigma(dt) = \mathbb{1}_{(0, \infty)}(t) t^{-1} e^{-t} dt$.

The kernel $K(x,y)$ is said to be totally positive on $X \times Y$, where $X, Y \subseteq \mathbb{R}$, if $$ K\pmatrix{x_1&x_2&\cdots&x_n\\y_1&y_2&\cdots&y_n} := \det \left|\matrix{K(x_1,y_1)&K(x_1,y_2)&\cdots&K(x_1,y_n)\\K(x_2,y_1)&K(x_2,y_2)&\cdots&K(x_2,y_n)\\\vdots&\vdots&&\vdots\\K(x_n,y_1)&K(x_n,y_2)&\cdots&K(x_n,y_n)\\}\right| \geqslant 0 $$ whenever $x_1 < x_2 < \ldots < x_n$ and $y_1 < y_2 < \ldots < y_n$ (and, of course, $x_1, x_2, \ldots, x_n \in X$, $y_1, y_2, \ldots, y_n \in Y$). It is strictly totally positive if strict inequality holds. A standard reference for totally positive kernels is Karlin's book Total positivity (Stanford, 1968). Our goal is thus to prove that the kernel $G(\mu,\nu) = \Gamma(\mu + \nu)$ is strictly totally positive.

It is known that the kernel $e^{x y}$ is strictly totally positive on $\mathbb{R} \times \mathbb{R}$; see, for example, Example (i) in Section 2.1 of Karlin's book. Substituting $t = e^y$, we see that $K(x, t) = t^x$ and $\check{K}(t, x) = t^x$ are strictly totally positive on $\mathbb{R} \times (0, \infty)$ and $(0, \infty) \times \mathbb{R}$, respectively.

Define $\sigma(dt) = t^{-1} e^{-t} dt$ on $(0, \infty)$. Observe that $$ G(\mu,\nu) = \Gamma(\mu + \nu) = \int_0^\infty t^{\mu + \nu - 1} e^{-t} dt = \int_0^\infty K(\mu, t) \check{K}(t, \nu) \sigma(dt) . $$ The basic composition formula (as it is called by Karlin, see (2.5) in Section 1.2 in his book; Karlin's reference for this formula is problem 68 in Pólya and Szegő, Aufgaben und Lehrsdtze aus der Analysis, vol. 1) tells us that $$ \begin{aligned} & G\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\\nu_1&\nu_2&\cdots&\nu_n} = \idotsint\limits_{0<t_1 < t_2 < \ldots < t_n} K\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\t_1&t_2&\cdots&t_n} \times \\ & \hspace{10em} \times \check{K}\pmatrix{t_1&t_2&\cdots&t_n\\\nu_1&\nu_2&\cdots&\nu_n} \sigma(dt_1) \sigma(dt_2) \ldots \sigma(dt_n) . \end{aligned} $$ The right-hand side is clearly positive, and our claim is proved.

The above argument is (essentially) contained in Karlin's book, when he proves that moment sequences generate totally positive kernels, see formula (2.10) in Section 1.2 of his book. He is only concerned with integer moments, but the argument carries over with no modifications to arbitrary moments.

This is merely a reformulation of Abdelmalek Abdesselam's answer, in a somewhat different language and with different references. It should be a comment to that answer, but it is unfortunately too long. Long story short: see Karlin, Total positivity, formula (2.10) in Section 1.2, with $u(t) = t$ and $\sigma(dt) = \mathbb{1}_{(0, \infty)}(t) t^{-1} e^{-t} dt$.

The kernel $K(x,y)$ is said to be totally positive on $X \times Y$, where $X, Y \subseteq \mathbb{R}$, if $$ K\pmatrix{x_1&x_2&\cdots&x_n\\y_1&y_2&\cdots&y_n} := \det \left|\matrix{K(x_1,y_1)&K(x_1,y_2)&\cdots&K(x_1,y_n)\\K(x_2,y_1)&K(x_2,y_2)&\cdots&K(x_2,y_n)\\\vdots&\vdots&&\vdots\\K(x_n,y_1)&K(x_n,y_2)&\cdots&K(x_n,y_n)\\}\right| \geqslant 0 $$ whenever $x_1 < x_2 < \ldots < x_n$ and $y_1 < y_2 < \ldots < y_n$ (and, of course, $x_1, x_2, \ldots, x_n \in X$, $y_1, y_2, \ldots, y_n \in Y$). It is strictly totally positive if strict inequality holds. A standard reference for totally positive kernels is Karlin's book Total positivity (Stanford, 1968). Our goal is thus to prove that the kernel $G(\mu,\nu) = \Gamma(\mu + \nu)$ is strictly totally positive.

It is known that the kernel $e^{x y}$ is strictly totally positive on $\mathbb{R} \times \mathbb{R}$; see, for example, Example (i) in Section 2.1 of Karlin's book. Substituting $t = e^y$, we see that $K(x, t) = t^x$ and $\check{K}(t, x) = t^x$ are strictly totally positive on $\mathbb{R} \times (0, \infty)$ and $(0, \infty) \times \mathbb{R}$, respectively.

Define $\sigma(dt) = t^{-1} e^{-t} dt$ on $(0, \infty)$. Observe that $$ G(\mu,\nu) = \Gamma(\mu + \nu) = \int_0^\infty t^{\mu + \nu - 1} e^{-t} dt = \int_0^\infty K(\mu, t) \check{K}(t, \nu) \sigma(dt) . $$ The basic composition formula (as it is called by Karlin, see (2.5) in Section 1.2 in his book; Karlin's reference for this formula is problem 68 in Pólya and Szegő, Aufgaben und Lehrsdtze aus der Analysis, vol. 1) tells us that $$ \begin{aligned} & G\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\\nu_1&\nu_2&\cdots&\nu_n} = \idotsint\limits_{0<t_1 < t_2 < \ldots < t_n} K\pmatrix{\mu_1&\mu_2&\cdots&\mu_n\\t_1&t_2&\cdots&t_n} \times \\ & \hspace{10em} \times \check{K}\pmatrix{t_1&t_2&\cdots&t_n\\\nu_1&\nu_2&\cdots&\nu_n} \sigma(dt_1) \sigma(dt_2) \ldots \sigma(dt_n) . \end{aligned} $$ The right-hand side is clearly positive, and our claim is proved.

The above argument is (essentially) contained in Karlin's book, when he proves that moment sequences generate totally positive kernels, see formula (2.10) in Section 1.2 of his book. He is only concerned with integer moments, but the argument carries over with no modifications to arbitrary moments.