$\Gamma(z)$ is only defined by the integral for $\mathfrak{Re}(z)>0$ because otherwise the integral diverges at the lower bound.
The modified function, for any fixed $n > 0$, is convergent for any $z \in \mathbb{C}$:
$$
g_n(z) = \int_{1/n}^\infty t^z \mathrm{e}^{-t} \frac{\mathrm{d} t}{t}
$$
However, as $n$ grows, the limit would diverge for $\mathfrak{Re}(z)<0$.
$g_n(z)$ is holomorphic by Looman-Menchoff's theorem. Indeed $g_n(z)$ is uniformly convergent, and hence differentiation and integration operations can be exchanged, allowing to conclude that Cauchy-Riemann equations are verified by $g_n(z)$.
Suppose $m = n + \alpha$ where $0 \le \alpha < 1$, so that $\alpha$ is the fractional part of $m$. Taking logarithms, the inequality becomes
$$ \sum_{k=1}^n \log (k + \alpha) + \log \Gamma (1+\alpha) <
\log 2 + (n+\alpha) \log (n+\alpha) + (n+\alpha) \log \frac{3}{5} $$
Now using Riemann sums, we have
$$ \sum_{k=1}^n \log (k + \alpha) < \int_1^{n+1} \log(x+\alpha) dx$$
which is
$$ (n+1+\alpha) \log (n+1+\alpha) - n - 1 - (1+\alpha) \log(1+\alpha) + 1.$$
This means we need to show that
$$ (n+1+\alpha) \log (n+1+\alpha) - n - (1+\alpha) \log(1+\alpha) + \log \Gamma (1+\alpha) < \log 2 + (n+\alpha) \log (n+\alpha) + (n+\alpha) \log \frac{3}{5} $$
Rearranging terms, this is
$$ (n+1+\alpha) \log (n+1+\alpha) - (n+\alpha) \log (n+\alpha) <
\log 2 - \log \Gamma (1+\alpha) + (n+\alpha) \log \frac{3}{5} + n + (1+\alpha) \log(1+\alpha)$$
Now for the LHS we have
$$ \log (n+1+\alpha) + \log \left( 1 + \frac{1}{n+\alpha}\right)^{n+\alpha} <
\log (n+1+\alpha) + 1$$
because $ \log \left( 1 + \frac{1}{x}\right)^x < 1$ by a trivial calculation.
On the RHS we have
$$ n \log\frac{3e}{5} + \log 2 - \log \Gamma (1+\alpha) + \alpha \log \frac{3}{5} + (1+\alpha) \log(1+\alpha) > \log (n+1+\alpha) + 1$$
for all $n > n_0$ for some $n_0$ because the coefficient $\log\frac{3e}{5}$ on $n$ is positive and the remaining terms are bounded by a constant. This concludes the proof. Note that the proof also goes through with a factor of $\frac{2}{5}$, just barely, and requiring $n_0 = 22.$ The original post has $n_0 = 1.$
I'm not sure of all the details but I hope it's a start.
Best Answer
The Bohr–Mollerup theorem shows that the gamma function is the only function that satisfies the properties
The condition of log-convexity is particularly important when one wants to prove various inequalities for the gamma function.
By the way, the gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=-n$, $n=0,1,2\dots$. There is a whole family of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{m=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\quad k\in\mathbb Z, $$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.
Edit: corrected index in the product.