I'll post an answer to spell out all the details.
You have $$G(ω)=\frac{1}{N}E\left[{\rm Tr}\frac{1}{Iω−J}\right]=\frac{1}{N}E\left[\sum_\lambda\frac{1}{ω−\lambda}\right]$$
This can be written as $$G(ω)=\frac{1}{N}E\left[\int d^2z \sum_\lambda\frac{\delta(z-\lambda)}{ω−z}\right],$$ where $\delta$ is the delta-function/distribution in the complex plane.
Now define $$\rho(z)=\frac{1}{N}E\left[\sum_\lambda\delta(z-\lambda)\right]$$, which is the spectral density. Then it follows that $$G(\omega)=\int d^2z \frac{\rho(z)}{\omega-z}=$$
Or maybe you mean it the other way around. If you start with $$\frac{1}{\omega+i\epsilon-\lambda}=\frac{\omega-i\epsilon-\lambda}{(\omega-\lambda)^2+\epsilon^2}$$, the imaginary part gives, in the limit of vanishing $\epsilon$ (as Beenakker mentioned), $$\lim_{\epsilon\to 0}{\rm Im}E\left[\sum_\lambda\frac{1}{\omega+i\epsilon-\lambda}\right]=\frac{1}{\pi}E\left[\sum_\lambda\delta(\omega-\lambda)\right]$$, which means that $\lim_{\epsilon\to 0}{\rm Im}G(\omega+i\epsilon)=\rho(\omega)N/\pi$.