Your first question can be answered with a standard "differentiation under the integral sign" proof:
Fix $z\in\Omega$ and let $(z_n)$ be an arbitrary sequence in $\Omega$ such that $z_n\to z$ and $z_n\neq z$ for all $n$. As $\mathbb{C}\setminus\Omega$ is closed and disjoint from the compact set $\{z_n\}_{n=1}^\infty\cup\{z\}\subset\Omega$, there is a fixed $\delta>0$ such that $|w-z|\geq\delta$ and $|w-z_n|\geq\delta$ for all $n$ and all $w\in\mathbb{C}\setminus\Omega$. Since $f(t)\in\mathbb{C}\setminus\Omega$ for all $t\in[a,b]$, it follows that $$\left|\frac{g(t)}{(f(t)-z_n)(f(t)-z)}\right|\leq\frac{|g(t)|}{\delta^2}\quad\text{for all }t\in[a,b].$$ Since $|g|/\delta^2$ is integrable on $[a,b]$, the above bound allows us to apply Lebesgue's Dominated Convergence Theorem to the following limit:
\begin{align}\lim_{n\to\infty}\frac{h(z_n)-h(z)}{z_n-z}&=\lim_{n\to\infty}\int_a^b\frac{g(t)}{(f(t)-z_n)(f(t)-z)}\,dt\\
&=\int_a^b\lim_{n\to\infty}\frac{g(t)}{(f(t)-z_n)(f(t)-z)}\,dt=\int_a^b\frac{g(t)}{(f(t)-z)^2}\,dt.
\end{align}
The value of this limit is independent of our choice of $(z_n)$, so we conclude that $$h'(z)=\int_a^b\frac{g(t)}{(f(t)-z)^2}\,dt.$$ In particular, $h'(z)$ exists for each $z\in\Omega$.
For your second question, note that $I(z)=h(e^z)$ where $$h(w)=\int_0^1\frac{\sin(t)}{t^2-w}\,dt.$$ After verifying that the domain of $h$ is $\mathbb{C}\setminus[0,1]$, apply the above result with $a=0$, $b=1$, $f(t)=t^2$, and $g(t)=\sin(t)$ to see that $h$ is analytic on $\mathbb{C}\setminus[0,1]$ and note that the preimage of $\mathbb{C}\setminus[0,1]$ along the exponential map $z\mapsto e^z$ is the connected open region $$\Omega=\mathbb{C}\setminus\{z\in\mathbb{C}:\Re(z)\leq 0\text{ and }\Im(z)\in 2\pi\mathbb{Z}\}.$$ Since the exponential map is analytic, it follows that $I$ is analytic on $\Omega$ and undefined elsewhere.
