The value of the integral will sum the residues at the poles that are interior to $A$. For $\omega$ in the restricted region, the denominator has two distinct roots, so there are either $0,1$, or $2$ simple poles within $A$ and its interior.
The roots of $z^2+2z-\omega$ are $-1\pm\sqrt{1+\omega}$. Let $U=\{\omega: |\omega|<3/4\}$. Then $\sqrt{1+U}$ is two disjoint regions in $\mathbb{C}$, one with positive real part and one with negative real part. So of the two roots of $z^2+2z-\omega$, one is to the left of $-1$ and exterior to $A$.
Certainly, there are values of $\omega$ for which the integrand has a pole at $c=-1+\sqrt{1+\omega}$ within $A$. The residue at a simple pole $c$ is $\begin{aligned} \lim_{z\to c}(z-c)\frac{z(z+1)}{z^2+2z-\omega}&=\lim_{z\to c}(z-c)\frac{z(z+1)}{(z-c)(z+\omega/c)}\\ &=\frac{c(c+1)}{c+\omega/c}\\ &=\frac{c(c^2+c)}{c^2+\omega}\\ &=\frac{c(\omega-c)}{2(\omega-c)}\\ &=\frac{(-1+\sqrt{1+\omega})(\omega+1-\sqrt{1+\omega})}{2(\omega+1-\sqrt{1+\omega})}\\ \end{aligned}$
Define $g(\omega)$ to be this last expression.
So $f(w)=g(w)$ if $\omega$ provides one pole within $A$. If the claim is to be true, then $f(w)=g(w)$ for all $w$ in $U$. The only way that the claim could fail to be true is if for some $\omega_0\in U$, the integrand has no poles within $A$ (making $f(\omega_0)=0$). (Although this would not be a problem if $\lim{\omega\to\omega_0}g(\omega)$ equals $0$ too.)
As noted in the comments, at $\omega=0$ there are no interior poles, but that's OK since the $g(\omega)$ approaches $0$ as $\omega\to0$.
It remains to show that when $\omega\in U\setminus\{0\}$, then $|-1+\sqrt{1+\omega}|<1/2$, and therefore there will be one pole interior to $A$. Let $V=\{c: |c|<1/2\}$. We are trying to show that $k(U)\subset V$, where $k(\omega)=-1+\sqrt{1+\omega}$. Since $k$ is one-to-one on $U$ and $V$, then this is equivalent to showing $U\subset k^{-1}(V)=\{c^2+2c:|c|<1/2\}$. My argument is that the boundary curve for $U$ ($z=\frac{3}{4}e^{it})$ and the boundary curve for $k^{-1}(V)$ ($z=\frac{1}{4}e^{2it}+e^{it}$) intersect tangentially at $t=\pi$ (corresponding to $z=-3/4$) and do not intersect anywhere else. Comparing $z$-values where $t=0$ reveals that $U$'s boundary is the interior curve.