TL;DR: Yes, $\Omega_{AB}(\rho)$ can be simplified. It turns out that it's one $n$th of the projector onto the ray spanned by the vectorization of $\rho^{1/2}$, see $(3)$ below.
First, observe that $\Omega_{AB}(\rho)$ is the outer product
$$
\Omega_{AB}(\rho) = |\psi_{AB}\rangle\langle\psi_{AB}|\tag1
$$
where $|\psi_{AB}\rangle = (\rho_A^{1/2}\otimes I_B)|\Psi^+\rangle$ which has norm $\frac{1}{\sqrt{n}}$. But
$$
\langle i_A|\langle j_B|\psi_{AB}\rangle = \frac{1}{\sqrt{n}}\sum_{k=1}^n\langle i_A|\rho_A^{1/2}|k_A\rangle\langle j_B|k_B\rangle=\frac{1}{\sqrt{n}}\langle i_A|\rho_A^{1/2}|j_A\rangle\tag2
$$
so $|\psi_{AB}\rangle$ is a scalar multiple of the vectorization of $\rho_A^{1/2}$. Thus, denoting$^1$ the vectorization of matrix $M_A\in L(\mathcal{H}_A)$ by $|M_{AB}\rangle\in\mathcal{H}_A\otimes\mathcal{H}_B$, we can rewrite $(1)$ as
$$
\Omega_{AB}(\rho) = \frac{1}{n}|\rho_{AB}^{1/2}\rangle\langle\rho_{AB}^{1/2}|\tag3
$$
where $|\rho_{AB}^{1/2}\rangle$ turns out to be normalized$^2$. Note that $\Omega_{AB}(\rho)$ is not actually a state, since $\mathrm{tr}(\Omega_{AB}(\rho))=\frac{1}{n}$.
That said, $\Omega_{AB}(\rho)$ may not necessarily be the most effective choice of a subexpression to isolate for simplification. See Rammus's comment for an alternative.
$^1$ Note that vectorization turns an "input" index into an "output" index. We are choosing to place the extra output index on the second system ($B$). This choice requires that $\mathcal{H}_A$ and $\mathcal{H}_B$ be isomorphic and is justified by the form of $|\Psi^+\rangle$. This accounts for the label change from $M_A$ to $|M_{AB}\rangle$.
$^2$ Perhaps the simplest way to see this is to note that the dot product $\langle M|N\rangle$ of vectorizations $|M\rangle$ and $|N\rangle$ is the Hilbert-Schmidt inner product $\mathrm{tr}(M^\dagger N)$ of $M$ and $N$.
