Yes, the almost partition exists. Instead of letting $\mu(E)\geq\frac{\mu(\Omega)}{2}$, I let $\mu(E)\in(0,\mu(\Omega))$ be arbitrary and proved that you can divide $\Omega$ into $N$ open simply connected districts $S',S_2,\dots,S_N$ of measure $\frac{\mu(\Omega)}{N}$ such that $\mu(S'\cap E)<\frac{\mu(E)}{N}<\mu(S_2\cap E)=\dots=\mu(S_N\cap E)$.
Lemma: Let $\mu$ be a finite measure on the open unit disk $D^2\subseteq\mathbb{R}^2$ equivalent to Lebesgue measure, let $E\subseteq D^2$ be a measurable set and let $n\in\mathbb{N}$. Then we can find a disk sector $S\subseteq D^2$ with $\mu(S)=\frac{\mu(D^2)}{n}$ and $\mu(S\cap E)=\frac{\mu(E)}{n}$.
Proof of the lemma: Let $f:[0,2\pi]$ be defined by $f(x)=\mu(\{z\in D^2;\text{arg}(z)\in(0,x)\})$. Then $f$ is a continuous function with $f(0)=0$ and $f(2\pi)=\mu(D^2)$. Now for each $t\in\left[0,\mu(D^2)-\frac{\mu(D^2)}{n}\right]$ let $S(t)$ be the sector $\left\{z\in D^2;\text{arg}(z)\in\left(f^{-1}(t),f^{-1}\left(t+\frac{\mu(D^2)}{n}\right)\right)\right\}$, so that $\mu(S(t))=\frac{\mu(D^2)}{n}$ for all $t$.
Note that $\mu(E\cap S(t))$ varies continuously, and at some point it has to be exactly $\frac{\mu(E)}{n}$: this is thanks to the intermediate value theorem and the fact that $\mu(E\cap S(0))+\mu(E\cap S(\frac{\mu(D^2)}{n}))+\dots+\mu(E\cap S(\frac{(n-1)\mu(D^2)}{n}))=\mu(E)$, due to $S(0),S(\frac{\mu(D^2)}{n})),\dots,S(\frac{(n-1)\mu(D^2)}{n}))$ being an almost partition of the disk. $\square$
Now let's return to the original problem. We can suppose that $\Omega$ is a disk and $\mu$ is some measure in $\Omega$ equivalent to Lebesgue measure, thanks to the Riemann mapping theorem. By the lemma there is a sector $S\subseteq\Omega$ with $\mu(S)=\frac{\mu(\Omega)}{N}$ and $\mu(S\cap E)=\frac{\mu(E)}{N}$. We can transform this sector a bit into a set $S'$ such that $\mu(S')=\frac{\mu(D^2)}{N}$ and $\mu(S'\cap E)<\frac{\mu(E)}{N}$: to do it, let $p\in S$ and $q\in D^2\setminus\overline{S}$ be such that $E$ has density $1$ in $p$ and $0$ in $q$. Now take very small balls $B_p,B_q$ around $p,q$, remove $B_p$ from $S$ and add $B_q$ to $S$. We can ensure that $S'$ and $D^2\setminus\overline{S'}$ are simply connected by joining $B_p$ to $D^2\setminus\overline{S}$ and $B_q$ to $S$ by small neighborhoods of curves (both neighborhoods can be taken to have the same area, so that $\mu(S')$ remains $\frac{\mu(D^2)}{N}$, and they can have area as small as we want, so that the inequality $\mu(S'\cap E)<\frac{\mu(E)}{N}$ is still satisfied).
The set $S'$ will be the only district where out party loses the majority vote. Let $\Omega_1=D^2\setminus\overline{S'}$, which is simply connected. Note that $\mu(S')=\frac{\mu(X)}{N}$, and we have $\frac{\mu(E\cap\Omega_1)}{\mu(\Omega_1)}>\frac{\mu(E)}{\mu(X)}$. By the lemma we can find an open simply connected subset $S_2$ of $\Omega_1$ such that $\mu(S_2)=\frac{\mu(\Omega)}{N}=\frac{\mu(\Omega_1)}{N-1}$ and $\frac{\mu(E\cap S_2)}{\mu(S_2)}=\frac{\mu(E\cap\Omega_1)}{\mu(\Omega_1)}$, and so that $\Omega_2=\Omega_1\setminus\overline{S_2}$ is simply connected. But then we also have $\frac{\mu(E\cap\Omega_2)}{\mu(\Omega_2)}=\frac{\mu(E\cap\Omega_1)}{\mu(\Omega_1)}$, so we can create $S_3$ using the lemma, etc.
At the end we are left with a sequence of districts $S',S_2,\dots,S_N=\Omega_{N-1}$ of measure $\frac{\mu(\Omega)}{N}$ such that
$$\frac{\mu(E\cap S')}{\mu(S')}<\frac{\mu(E)}{\mu(\Omega)}<\frac{\mu(E\cap\Omega_1)}{\mu(\Omega_1)}=\frac{\mu(E\cap S_2)}{\mu(S_2)}=\dots=\frac{\mu(E\cap S_N)}{\mu(S_N)},$$
which after multiplying by $\frac{\mu(\Omega)}{N}$ becomes what we wanted:
$\mu(S'\cap E)<\frac{\mu(E)}{N}<\mu(S_2\cap E)=\dots=\mu(S_N\cap E)$.
