A bit late to the game, but here is an answer:
1: To show it is finite: Write $e^{-x}=e^{-x/2}\cdot e^{-x/2}$. For every $y>0$ there exists $N$ such that $e^{-x/2}x^{y-1}<1$ when $x>N$ so that $$\int_N^\infty e^{-x} x^{y-1}dx\leq \int_N^\infty e^{-x/2} <\infty.$$ For $x$ between $0,1$ compare to $\int x^{y-1}dx$ which converges whenever $y-1>-1$, so for all $y$. Lastly bounding $\int_1^N e^{-x}x^{y-1}dx$ I leave to you.
2: For this part, what you have done so far is good. All we need to do is switch the last integral with the limit. To do this, notice that in some neighborhood of radius $\delta$ around $y_0$ we can bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ by similar methods as in 1. Then, applying the Dominated Convergence Theorem to that neighborhood, we can switch the limit and the integral, solving the problem.
Remark: For that last part, notice we cannot bound $$\int_0^\infty e^{-x} x^{y-1}dx$$ uniformely for all $y\in(0,\infty)$ since the function blows up at $0$ and at $\infty$. But fortunately, we can bound it uniformely for all $y$ in any compact subset of $(0,\infty)$, allowing us to use the DCT.
As you have said, $\displaystyle \int_0^{\infty}t^{z-1}e^{-t}dt$ does not even make sense on the left half plane.
However, if we define $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ on the left half plane, (and this is not equal to the integral above on the left half plane), this is consistent with the fact that $\Gamma(z) = \displaystyle \int_0^{\infty}t^{z-1}e^{-t}dt$ on the right half plane and also provides us a possible extension on the left half plane except at negative integers.
Further, this has to be the only analytic continuation since $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ is analytic everywhere except at negative integers and matches with the integral on the right half plane and hence by uniqueness of analytic continuation this is the only possible extension.
EDIT
To see that the extension, is in fact analytic except at negative integers, you proceed strip by strip. On the right half plane, you have $\Gamma(z)$ defined by the integral, $\displaystyle \int_0^{\infty} t^{z-1} \exp(-t) dt$. You can verify that the integral satisfies the functional equation and is in fact analytic.
We will now extend the function to the strip corresponding to $(-1,0)$ i.e. the strip $\text{Re}(z) \in (-1,0)$ based on the functional equation.
Define $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ on the strip $\text{Re}(z) \in (-1,0)$.
First note that $\Gamma(z+1)$ is analytic on the strip $\text{Re}(z) \in (-1,0)$ since $z+1 \in$ right half plane. Also, $\dfrac1z$ is analytic on the strip $\text{Re}(z) \in (-1,0)$. Hence, $\Gamma(z) = \dfrac{\Gamma(z+1)}{z}$ is analytic on the strip $\text{Re}(z) \in (-1,0)$. Hence, now we have extended the definition of the $\Gamma$ function to the region $\text{Re}(z) > -1$.
Now repeat the same argument for the strip $\text{Re}(z) \in (-2,-1)$, since we now have that the $\Gamma$-function is analytic on the region $\text{Re}(z) > -1$. And so on...
Best Answer
$\Gamma(z)$ 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 $g_n(z)$.