The first thing to point out is that there are two equivalent ways to describe a quantum statistical distribution: the density matrix, and a probability distribution on the results of measurements of a "complete" set of observables. (It is remarkable that one should think of givin such a probability distribution on the classical phase space of the quantum system, but let's not go into this here. It is remarkable since it seems to go against the whole principle of quantum mechanics to assign a probability to a point with simultaneously assigned values of both position and momentum. But it works somehow anyway.)
Landau explains this equivalence between $w$ as a probability distribution and $w$ as a density matrix. In his formula he wants to point out the analogy with classical mechanics and for that reason interprets $w$ as a probability distribution.
The microcanonical ensemble consists only of states with a given fixed energy $E_0$: any other value of the energy has a zero probability of resulting from a measurement. But of course there are other commuting observables, and they will take on possibly different values when measured, so $w$ is a distribution on those values. But it is, by definition, somehow "constant". Other statistical distributions on this same set of states would be possible, but would not be the microcanonical one.
The second edition of Landau--Lifschitz explicitly says that $dw$ is the probability of finding the system in one of the states that "belongs" to one of the energy levels infinitesimally close to $E_0$. He also says explicitly that
$d\Gamma$ is the number of quantum states with energies in that same infinitesimal range around $E_0$. It is very hard to make rigorous sense out of this, although it has been done by other authors. That is why the density matrix is often preferred.
In fact the multiplication by the delta function means all states with a non-zero probability have the same energy. So $w$ on that ensemble is not a function of $E_0$, since $E_0$ is a parameter. But if you look at that formula, there are no other variables except $E$. So $w$ is constant on that ensemble, i.e. every state has the same probability. So we don't really need a formula at all: the constant $d\Gamma$ doesn't affect relative probabilities, and if you normalise this to make them genuine probabilities you just divide by $d\Gamma$.... Landau was a great physicist but he was not a very logical writer. Nor a very clear one. And throwing a junior co-author into the mix just makes things worse...
Varying the parameter and integrating $w$ over $E_0$ makes no real sense for the micro-canonical distribution.
The point is if you wish to calculate the entropy of the microcanonical distribution, as a function of the parameter $E_0$ (which is what Landau
does in the very next section). Because the normalised density matrix has to have trace one, and be diagonal, and since every probability is the same, the value of the diagonal elements is just the reciprocal of the dimension of the space of states with that energy, which is the number or elements in a basis, which is what Landau really means by "the number of quantum states that 'belong' to an energy in that infinitesimal range." (Literally, of course, the number is uncountably infinite, not finite.)
I never recommend reading Landau unless you already understand the basic concepts. Landau is the kind of chess grandmaster who can be champion of the world but could not explain to you in writing how a knight moves.
A point in the phase space corresponds to a microstate of a system of interest.
The ensemble is the manifold of different possible microstates compatible with some external constraint.
To evaluate an average of some observable over a subset of microstates, we need a measure over the microstates, and the measure of the totality of microstates (the whole accessible phase space in that ensemble) must be finite. Therefore we can always get a probability measure. The celebrated Radon-Nykodym theorem ensures that all the different measures can be recast in the form of suitable probability densities.
Therefore, nothing happens to the ensemble when we integrate in the phase space. We just evaluate phase space averages that can be used in place of time averages in the case of ergodic systems.
An ensemble is characterized by its probability density. Generally, it is a function of the phase space point, the parameters required to represent a macrostate, and time. It turns out that time-independent densities imply time-independent averages. Therefore time independent densities are necessary to describe equilibrium macrostates.
However, this does not imply that there is no microscopic dynamics at equilibrium inside a single system. The individual microstates have their dynamics. But the resulting probability density does not change at equilibrium.
Best Answer
There are a few concepts that should be better focused, to formulate this question precisely.
An ensemble of Classical Statistical Mechanics is the set of all possible configurations in phase space, each configuration being characterized by the set of its Hamiltonian coordinates $q=(q_1,q_2,\dots,q_N)$ and $p=(p_1,p_2,\dots,p_N)$. Therefore, there is nothing like the probability of an ensemble being in an element of phase space. Instead, we can safely speak about the probability of a system of the ensemble being in a volume of the phase space. When such a volume is so small that the variations of the probability density over the volume are negligible, we can say that the probability of that microscopic state is ${\mathrm dP}=\rho(q,p){\mathrm dq}{\mathrm dp}$.
If $\rho(q,p)$ is a constant over the hypersurface $H(q,p)=E$ (where $H$ is the Hamiltonian, and $E$ a possible value of the energy), all the subsets of the phase space on such a hypersurface with the same volume ${\mathrm dq}{\mathrm dp}$ have the same probability.
This fact implies that each microstate is as probable as any other. However, a set of microstates may be overwhelming more probable than others. In particular, the collection of microstates such that all the particles occupy only half of the volume has a negligible probability compared to the set where there is almost the same number of particles in the two half-volumes.