I'll suppose both $E_0$ and $E_1$ are bounded. Let $\Gamma_0$ and $\Gamma_1$ be disjoint positively-oriented simple closed contours in $\Omega$ enclosing $E_0$ and $E_1$ respectively, and $\Gamma_2$ a large positively-oriented circle enclosing both $\Gamma_0$ and $\Gamma_1$. Let $\Omega_1$ be the region inside $\Gamma_2$ but outside $\Gamma_0$ and $\Gamma_1$. Then for $z \in \Omega_1$ we have by Cauchy's integral formula, $ f(z) = \frac{1}{2\pi i} \left( \int_{\Gamma_2} \frac{f(\zeta)\ d\zeta}{\zeta - z} - \int_{\Gamma_0} \frac{f(\zeta)\ d\zeta}{\zeta - z} - \int_{\Gamma_1} \frac{f(\zeta)\ d\zeta}{\zeta - z} \right)$
If you're not familiar with this version of Cauchy's formula, you can draw thin "corridors" connecting $-\Gamma_0$, $-\Gamma_1$ and $\Gamma_2$ into a single closed contour enclosing $z$.
If $f_k(z) = \frac{1}{2\pi i} \int_{\Gamma_k} \frac{f(\zeta)\ d\zeta}{\zeta - z}$ this says $f(z) = f_2(z) - f_0(z) - f_1(z)$, where $f_2(z)$ is analytic everywhere inside $\Gamma_2$, $f_0(z)$ is analytic everywhere outside $\Gamma_0$, and $f_1(z)$ is analytic everywhere outside $\Gamma_1$. Moreover, the values of $f_k(z)$ don't depend on the choice of contours, as long as $z$ is inside $\Gamma_2$ and outside $\Gamma_0$ and $\Gamma_1$. By making $\Gamma_2$ sufficiently large and $\Gamma_0$ and $\Gamma_1$ sufficiently close to $E_0$ and $E_1$, any point in $\Omega$ can be included. So we actually have $f(z) = f_2(z) - f_0(z) - f_1(z)$ everywhere in $\Omega$, with $f_2(z)$ entire, $f_0(z)$ analytic outside $E_0$ and $f_1(z)$ analytic outside $E_1$.