I will address your second question: "one has to prove that the classification diagram functor is sent under this Quillen equivalence to something weakly equivalent to the coherent nerve".
The answer is yes, this is true. First note that the simplicial category $M^\circ$ is equivalent as a simplicial category to $L^HM$, the hammock localization of M (at the weak equivalences). See here for references. So the answer to your second question (or is it a remark?) follows as a special case of this previous MO answer (which cites work of Barwick-Kan and Toen). So the classification diagram functor and the coherent nerve of $M^\circ$ are sent to equivalent things under the Quillen equivalence between quasicategories and complete Segal spaces (I prefer the term Rezk categories or Rezk $\infty$-categories).
I believe that combined with your other comments (notably using Julie Bergener's results) this is enough to answer your main question in the affirmative.
I'm going to deal with the case of $\mathcal E^1\subset \mathcal E^0$. Note that by composition of pullbacks, $\mathcal E^0 = \mathcal C_{/X}\times_{\mathcal P(S)}\mathcal P(S)_{j(s)/}$, while $\mathcal E^1= \mathcal C_{/X}\times_\mathcal C\{j(s)\}$.
Further, because $\mathcal C\to \mathcal P(S)$ is a full subcategory inclusion, you can rewrite $\mathcal E^0 = \mathcal C_{/X}\times_\mathcal C \mathcal C_{j(s)/}$
Now we write the obvious thing : the homotopy $\mathcal E^0\times\Delta^1\to \mathcal E^0$ should be given by the following thing:
- on the $\mathcal C_{j(s)/}$ factor, go $\mathcal E^0\times\Delta^1\to \mathcal C_{j(s)/}\times\Delta^1\to \mathcal C_{j(s)/}$ where the second map is the canonical homotopy witnessing that $\mathcal C_{j(s)/}$ deformation retracts onto $\{j(s)\}$;
- on the $\mathcal C$ factor, same as above but postcompose further with $\mathcal C_{j(s)/}\to \mathcal C$
- On the $\mathcal C_{/X}$ factor, you'll have to observe that there is a composition map $\mathcal C_{/X}\times_\mathcal C \mathcal C_{j(s)/}\to map(X,j(s))$. Indeed this is like $\{j(s)\}\times_{\hom(\{0\},\mathcal C)} \hom(\Lambda^2_1,\mathcal C)\times_{\hom(\{2\}, \mathcal C)} \{X\}$, and the forgetful map $\hom(\Delta^2,\mathcal C)\to \hom(\Lambda^2_1,\mathcal C)$ is an acyclic fibration, so in particular has a section. So you can use that section, and then evaluate on $\Delta^{\{0,2\}}$, and forget about the $\{j(s)\}\times_{\hom(\{0\},\mathcal C)}$ part to land in $\hom(\Delta^1,\mathcal C)\times_{\hom(\{2\},\mathcal C)}\{X\}= \mathcal C_{/X}$.
Here I'm being a bit sloppy between $\mathcal C_{/X}$ and $\hom(\Delta^1,\mathcal C)\times_{\hom(\{1\},\mathcal C)}\{X\}$ (I always forget whether they are literally isomorphic or just equivalent) but they are equivalent so it does not really matter.
So now, you have an equivalence ${}_{j(s)}\hom(\Lambda^2_1,\mathcal C)_X\to {}_{j(s)}\hom(\Delta^2,\mathcal C)_X$, and the point is now that there is a homotopy $\Delta^1\times\hom(\Delta^2,\mathcal C)\to \hom(\Delta^1,\mathcal C)$ that more or less witnesses the fact that there is a map from $0\to 2$ to $1\to 2$ in the triangle $\Delta^2$, and so if you string these together, you will get a homotopy $\Delta^1\times\mathcal E^0\to \mathcal C_{/X}$ that looks exactly like the natural transformation $(j(s)\to y\to X, j(s) = j(s))\to (y\to X, j(s)\to y)$ that you would write down $1$-categorically.
Now these three homotopies are compatible and so they do assemble as a map from $\mathcal E^0\times\Delta^1$ to the pullback, which is precisely $\mathcal E^0$. It is clear from the construction that it starts at the identity, and ends in $\mathcal E^1$.
Basically the key point here was the third homotopy: given what you wrote in your question about "composition", I'm guessing that this is what was missing. In particular, I would suggest trying to see if you can do the same kind of the thing for the case of $\mathcal E$, and replace the words "compose" in your $1$-categorical proof with something along the lines of what I did here. If that doesn't work, let me know and I'll try to modify my answer to incorporate that part as well.
Also note that, as Zhouhang pointed out in the comments, Kerodon has a different proof of this fact altogether, or rather it's arranged differently : it observes that the yoneda embedding is dense, and earlier proved that if $f : C\to D$ is dense, then the identity of $D$ was the left Kan extension of $f$ along itself (and also relates this to the condition that the "canonical diagram" be a colimit diagram). It feels like a better proof, so I would suggest looking at that too.
Best Answer
As shown by Dwyer–Kan (Function complexes in homotopical algebra, Proposition 4.8), for a simplicial model category $A$, the simplicial category $A^\circ$ is Dwyer–Kan equivalent to the simplicial category given by the hammock localization of $A$ with respect to the weak equivalences of $A$.
Applying this fact to the model category $A^C$ equipped with its projective or injective model structure, and using the fact that weak equivalences are the same for both model structures, we arrive at the desired result.