I was able to show that the integral formula for the gamma function
$$\Gamma(s) = \int_0^\infty x^{s-1}e^{-x}dx$$
is uniformly convergent for $s$ in any closed and bounded interval using the Weierstrass theorem. Now I would like to show that the convergence is not uniform for all $s > 0$ but am not sure how to proceed.
The integral representation converges for $Re s > 0$. As an improper integral, convergence is indeed uniform for (real) $s \in [\alpha,\beta]$. However, as you suspect, convergence is not uniform for $s \in (0,\infty)$. Problems arise due to singular behavior both at $x = 0$ when $s < 1$ and as $x \to \infty$ when $s > 1$.
To show non-uniform convergence consider
$$\Gamma(s) = \Phi(s) + \Psi(s),$$
where
$$\Phi(s) = \int_0^1 x^{s-1} e^{-x} \, dx, \\ \Psi(s) = \int_1^\infty x^{s-1} e^{-x} \, dx.$$
We can show that $\Psi(s)$ fails to converge uniformly for $s \in [\alpha, \infty).$ Note that for any sequence $(s_n)$ such that $s_n \to \infty$ with $s_n > 1 + n/\log n,$ we have
$$\left| \int_n^\infty x^{s_n-1} e^{-x} \, dx\right| > n^{s_n-1}\int_n^\infty e^{-x} \, dx = n^{s_n-1}e^{-n} > 1,$$
Thus,
$$\lim_{n \to \infty}\int_n^\infty x^{s_n-1} e^{-x} \, dx \neq 0,$$
and convergence is not uniform.
Similarly, we can show that $\Phi(s)$ fails to converge uniformly for $s \in (0, \beta).$ Note that for $s_n < 1$,
$$\left| \int_{1/2n}^{1/n} x^{s_n-1} e^{-x} \, dx\right| > e^{-1} \left(\frac{1}{n}\right)^{s_n-1}\frac{1}{2n} = \frac{1}{2en^{s_n}}.$$
Now take $s_n = 1/n$. In this case $s_n \to 0$ and $n^{s_n} = n^{1/n} \to 1$. For all $n$ sufficiently large, we have $n^{1/n} < 2$ and
$$\left| \int_{1/2n}^{1/n} x^{s_n-1} e^{-x} \, dx\right| > \frac{1}{4e}.$$
