The problem is that you cannot choose a domain and codomain for such a putative adjunction consistently and simultaneously. The statement that we have is that the category of presheaves on $C$ is the free cocomplete category on $C$ when $C$ is small. However, the forgetful functor from cocomplete categories to categories does not land in small categories-every cocomplete category which is not a preorder is large.
So, you might want an improved adjunction between large categories and cocomplete categories. However, the category of presheaves on a large category is even larger than large! What this means depends on your foundations. If we work with universes so that a small category is $U_1$-small, then presheaves on a small category are small with respect to the next biggest universe $U_2$. Now presheaves on a $U_2$-small category only have the appropriate universal property if we can them to be presheaves of $U_2$-small sets, and such presheaves are not $U_2$-small.
So the presheaves-forgetful functor pair cannot form an adjunction, because the desired left adjoint moves us up a universe every time we apply it. Thus in particular we cannot get around this straightforwardly by using universes.
There are a couple of partial solutions to this problem. The simplest is to regard the formation of the presheaf category as a relative left adjoint to the forgetful functor from cocomplete categories to (possibly) large categories. In other words, it behaves like a left adjoint, but is only partially defined. This is a rephrasing of the theorem you quote-the formation of presheaves behaves like a left adjoint when its input is small.
A more technical approach is to ask the question: even if the formation of presheaves cannot be left adjoint to the forgetful functor from cocomplete categories to large categories, does this forgetful functor have any left adjoint at all? In fact it does; for instance, it satisfies a 2-categorical version of the general adjoint functor theorem. This left adjoint sends a category $K$ to the subcategory of presheaves on $K$ formed by the colimits of small diagrams of representable presheaves. However, this category of small presheaves is not nearly as well behaved as the presheaf category. It needn't even be a topos in general.
So to summarize, this is a real issue which cannot be eradicated by any level of generous assumptions on the foundations. It's the go-to example of why size issues cannot be completely ignored in category theory.
Your first question doesn't really make sense, or rather it's not reasonable to expect it to be the identity : that would mean you have $u_!(X) = F(X)$ with a hard equal, given the definition of $u_!$, it's not reasonable.
However, you can indeed note that $u_!(X)$ has two isomorphisms with $F(X)$ when $X$ is representable : the one you constructed here, and the one given in your other question : you can ask whether they're the same (i.e. if you consider the second one to be an identification, then the second one is the identity with respect to that identification - I think that's what Cisinski means)
The answer is yes.
Indeed, note that for a represented presheaf $X= \hom(-,b)$, we have $\hom( u_!X,Y) \cong \hom(X, u^*Y) \cong u^*Y(b) = \hom(u(b), Y) = \hom(F(\hom(-,b)), Y)$
The string of isomorphisms up to $\hom(u(b), Y)$ yields the iso $u_!X \cong u(b)$ that you had defined in your previous question (by definition); and then you can note that since $(b,id_b)$ is terminal in $\int X$, $u(b) \to u_!X$ ($\mu^X_{(b,id_b)}$, the canonical inclusion) is an isomorphism, and it suffices to check that this is indeed the same as the one given by our string of isomorphisms.
But to check this, one only needs to remember where the adjunction $u_! \dashv u^*$ came from in the first place : precisely from the same type of canonical inclusion. Let's take $Y= u_!X$ and follow $id_{u_!X}$ : it goes to $a\mapsto (s\mapsto (u(a)= u_X(a,s) \overset{\mu^X_{(a,s)}}\to u_!X\to u_!X))$ so to $a\mapsto (s\mapsto (u(a)= u_X(a,s) \overset{\mu^X_{(a,s)}}\to u_!X))$;
then you evaluate that in $b,id_b$ so you get precisely $u(b) \overset{\mu^X_{(b,id_b)}}\to u_!X$.
So the two isomorphisms are the same in the case of a representable presheaf, which is the best we can get if we want $\eta$ to be "the identity on representable presheaves".
For your last question, this simply follows from any presheaf being a canonical colimit of representable presheaves : if you have two natural morphisms $h,k : F\to G$ between a colimit preserving functor $F$ and $G$ any functor on $\widehat{A}$ which agree on representables, then $h=k$.
Indeed, let $X$ be any presheaf, we want to check that $h_X = k_X$; for that it suffices to show that $h_X \circ F(\lambda^X_{(a,s)}) = k_X \circ F(\lambda^X_{(a,s)})$ for any $(a,s) \in \int X$ (by the definition of colimit, and because $F$ respects them)
But this follows because the LHS is just $G(\lambda^X_{(a,s)})\circ h_{\hom(-,a)}$ by naturality, and the RHS is $G(\lambda^X_{(a,s)})\circ k_{\hom(-,a)}$ for the same reason, but $h_{\hom(-,a)} = k_{\hom(-,a)}$ by assumption, so RHS = LHS, and so $h=k$.
Since $u_!$ preserves colimits, there is at most one natural iso that agrees with $\eta$ on representables : this is the uniqueness statement you wanted
Best Answer
I've modified your notation slightly. $x_0$ is the point instead of $x$, and $\psi$ is indexed by pairs $x,y$ in $D$.
To show universality, you want to suppose that for any set $S$ with a cocone of maps $\psi_{x,y}' : \newcommand\Hom{\operatorname{Hom}}\Hom(x_0,x)\to S$, you can construct a unique map $\alpha : F(x_0)\to S$.
Then given $y\in F(x_0)$, define $\alpha(y)=\psi_{x_0,y}'(\newcommand\id{\operatorname{id}}\id_{x_0})$.
Then we check that $\alpha \circ \psi_{x,y} = \psi'_{x,y}$.
Suppose $g\in \Hom(x_0,x)$, then $\psi_{x,y}(g) = F(g)(y)$.
Then $$\alpha(\psi_{x,y}(g)) = \alpha(F(g)(y))=\psi'_{x_0,F(g)(y)}(\id_{x_0}).$$
Observe that given $g\in \Hom(x_0,x)$, $g$ defines a morphism between the pairs $(x,y)$ and $(x_0,F(g)(y))$ by definition, so we must have $\psi'_{x_0,F(g)(y)} = \psi'_{x,y}\circ g_*$.
Thus $\alpha(\psi_{x,y}(g)) = \psi'_{x,y}(g_*\id_{x_0}) = \psi'_{x,y}(g)$, as desired.
As for uniqueness, suppose $\beta$ were another appropriate map $F(x_0)$ to $S$.
Then since $y = \id_{F(x_0)}y = F(\id_{x_0})y = \psi_{x_0,y}(\id_{x_0})$, for any $y\in F(x_0)$, we have $\beta(y) = \beta(\psi_{x_0,y}(\id_{x_0}))=\psi'_{x_0,y}(\id_{x_0}) = \alpha(y)$. Thus $\alpha$ is unique.