That definition of $\Omega$ only works if you have distinguishable particles and you can put as many as you want in any state. That gives you the Maxwell-Boltzmann distribution.
In quantum mechanics, particles of the same type are not distinguishable; e.g. there's no way to tell one electron from another. Moreover, for Fermions (like electrons), you can have at most one particle in a given state. Bosons (like photons) can have as many particles in one state as you please.
These give rise to two '$\Omega$'s, both different than the one you gave.
As for the partition function, it takes the same form, but you sum over different things. Define
$$Z=\sum_se^{-E_s\beta}$$
where the sum is over the possible microstates of the system. (A state is an energy level that a particle can occupy; e.g. the ground state of a hydrogen atome. A microstate is a configuration of all the particles in the system; some examples are given below.)
Say you have three particles and three states available to you. For Bosons, you have many different microstates available to you: all three particles in state one, one particle in each state, all three particles in state three, etc. So, your sum has many terms. For Fermions, you have only one option: one particle in each state, since there are three particles, three states, and only one particle can be in each state. So there is only one microstate available for Fermions in this system, and thus the sum only has one term.
Short version: the equation for the partition function stays the same, but the microstates you can sum over are very different.
The difference lies in the way we count the number of states of the system in quantum and classical cases.
The formulas you wrote are actually for the grand canonical partition functions for a single energy state, not for the whole system including all the energy states. The total grand canonical partition function is $$\mathcal{Z} = \sum_{all\ states}{e^{-\beta(E-N\mu)}} = \sum_{N=0}^\infty\sum_{\{E\}}{e^{-\beta(E-N\mu)}}$$
Now, if the particles are bosons, then the energy eigenstates are countable as $\{\epsilon_i\}$ and $\mathcal{Z}$ would be $$\mathcal{Z} = \sum_{\{n_i\}}e^{-\beta\sum_{i}{n_i(\epsilon_i-\mu)}} = \sum_{\{n_i\}}\prod_ie^{-\beta n_i(\epsilon_i-\mu)}=\prod_i \mathcal{Z_i}^{B-E}$$
where $n_i$ is the number of particles in $i$-th energy state, thus $\sum_i{n_i}=N$, and
$$\mathcal{Z_i}^{B-E} = \sum_{n=0}^\infty e^{-n\beta(\epsilon_i-\mu)}$$
Here, $\mathcal{Z_i}^{B-E}$ is the grand canonical partition function for one energy eigenstate with energy $\epsilon_i$ in Bose-Einstein statistics.
On the other hand, in classical regime the energy of a particle can take any energy. In this case, one point in the $6N$-dimensional phase space denotes one state of system. Therefore, the energy states are not countable as there is an infinite number of points in the phase space within any phase space volume. To count the states we take the "semi-classical" approach by taking the phase space volume of one state of the system to be $(2\pi\hbar)^{3N}$. We can then integrate over the whole phase space and divide the integral by this unit volume to get the number of states. However, as the particles are assumed to be indistinguishable, any permutation of the system configuration (the set of $\{\vec{x}_n,\ \vec{p}_n\}$) would actually be the same state of the system. Therefore, when we integrate over the whole phase space volume we overcount the total number of states by $N!$. That's why we need to divide the integral by Gibbs factor $N!$. For a system of non-interacting particles, the N-particle canonical partition function then can be written as $Z_N = \frac{Z_1^N}{N!}$ where $Z_1$ is the canonical partition function for one particle.
Now, the grand canonical partition function for a classical system would be
$$\begin{align}
\mathcal{Z}&=\sum_{N=0}^\infty{\int_0^\infty dE\ \Omega(E,N)e^{-\beta(E-\mu N)}}
=\sum_{N=0}^\infty e^{\beta\mu N} Z_N = \sum_{N=0}^\infty e^{\beta\mu N} \frac{Z_1^N}{N!}\\
\end{align}$$
If we want to derive the partition function $\mathcal{Z_i}$ for a single energy state similar to the Bose-Einstein statistics, we can assume the energy of a single particle to be discrete and countable as $\{\epsilon_i\}$. This can be achieved by dividing the one particle phase space into s of unit volume and assigning one representative energy to every unit volume section. Then, the grand canonical partition function is,
$$ \begin{align}
\mathcal{Z} &= \sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} (\sum_i e^{-\beta\epsilon_i})^N\\
&=\sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} \sum_{\{n_i\},\sum n_i = N} \frac{N!}{\prod_i n_i!} e^{-\beta\sum_i n_i\epsilon_i}\\
&=\sum_{\{n_i\}}\prod_i \frac{1}{n_i!} e^{-\beta n_i(\epsilon_i-\mu)}\\
&=\prod_i \sum_{n=0}^\infty \frac{1}{n!} e^{-\beta n(\epsilon_i-\mu)} = \prod_i \mathcal{Z_i}^{M-B}
\end{align}$$
This $\mathcal{Z_i}^{M-B}$ is the single energy state grand canonical partition function in Maxwell-Boltzmann statistics.
Maxwell-Boltzmann statistics is the classical limit for Bose-Einstein statistics. The condition for the system to be classical is the single state occupation number $\bar{n}$ to satisfy $\bar{n} \ll 1$, in other words, the total number of single particle states $M$ should satisfy $N \ll M$. As, the single particle partition function $Z_1$ is actually a weighted sum over all the states, $N\ll Z_1$ will satisfy $N \ll M$ for the system to be classical.
Best Answer
The factorial factor $1/N!$ is exact but does not apply to all statistics.
Consider a two level system and let us call $\xi_i$ the grand-canonical partition function for the energy level $i$ ($i=0$ or $1$) and $z=\mathrm e^{\beta\mu}$ the fugacity. Keep in mind that $\xi_i$ is the partition function for a given energy level.
Classical particles are independent, indistinguishable, $\xi_i$ is computed as a usual partition function for undistinguishable particles. Here we divide by the symmetry $p!$ because the particles are undistinguishable and therefore the order in which we could label them should not matter. $$\xi_i^{\text{classical}}=\sum_{p=0}^\infty \frac1{p!}z^p \mathrm e^{-p\beta E_i}=1+z\mathrm e^{-\beta E_i}+\frac{z^2}{2}\mathrm e^{-2\beta E_i}+\cdots=\exp\left(z\mathrm e^{-\beta E_i}\right).$$ For quantum particles, the sum runs over the number of particles which can occupy the energy level. There is no division by $p!$ because the quantum states already are (anti)-symmetric; for instance the quantum state where there is one fermion in each level is $\frac1{\sqrt2}\left(\left|01\right\rangle-\left|10\right\rangle\right)$. $$ \xi_i^{\text{fermions}}=\sum_{p=0}^1z^p\mathrm e^{-p\beta E_i}=1+z\mathrm e^{-\beta E_i}$$ $$ \xi_i^{\text{bosons}}=\sum_{p=0}^\infty z^p\mathrm e^{-p\beta E_i}=\frac1{1-z\mathrm e^{-\beta E_i}}$$ The grand-canonical partition functions are $$ \mathcal Z^{\text{classical}}=\xi_0^{\text{classical}}\xi_1^{\text{classical}}=\mathrm e^{z\left(1+\mathrm e^{-\beta E}\right)} =\mathrm e^{zZ_1}=1+zZ_1+\frac{z^2}2Z_1^2+\cdots$$ $$\mathcal Z^{\text{fermions}}=\xi_0^{\text{fermions}}\xi_1^{\text{fermions}}=(1+z)\left(1+z\mathrm e^{-\beta E}\right)=1+zZ_1+z^2\mathrm e^{-\beta E}$$ $$\mathcal Z^{\text{bosons}}=\xi_0^{\text{bosons}}\xi_1^{\text{bosons}}= \frac1{1-z}\frac1{1-z\mathrm e^{-\beta E}}=1+zZ_1-z^2\mathrm e^{-\beta E}+z^2Z_1^2+\cdots$$ with $Z_1=1+\mathrm e^{-\beta E}$ is the one-particle partition function.
Now looking at the coefficient $z^2$ in $\mathcal Z$ gives the two-particle canonical partition function $Z_2$. We have $$ Z_2^{\text{classical}}=\frac{1}{2}Z_1^2,\quad Z_2^{\text{fermions}}=\mathrm e^{-\beta E},\quad Z_2^{\text{bosons}}=1+\mathrm e^{-\beta E}+\mathrm e^{-2\beta E}.$$ In none of these case was it necessary to make an approximation. Undistinguishability in the quantum cases is for quantum states, not for particles.
Remark The difference between $Z_2^{\text{classical}}$ and $Z_2^{\text{bosons}}$ is evidence for the fact that the bosons and the classical particles have a difference. Indeed, classical particles have no correlations, which is expressed by the fact $Z_2\propto Z_1^2$, whereas bosons have correlations: compared to the uncorrelated classical particles, the relative weight of the states where the particles are both at the same energy level is larger: bosons prefer to be in the same state. (Fermions avoid being at the same energ level and classical particles don't care.)