Lemma 3.34: For $F,G,H$ presheaves in discrete categories/sets: $$\mathfrak{S}_F \times_{\mathfrak{S}_H} \mathfrak{S}_G \cong \mathfrak{S}_{F \times_H G}$$
Proof: The only 2-morphisms of categories fibered in sets are identities. (Ref: http://wwwf.imperial.ac.uk/~dg3215/other/stacks.pdf )
Question: I am not quite sure of the proof where they make use of 2-morphisms of categories fibered in sets are identities in proving the lemma.
Attempt: We want to show an equivalence of categories between $\mathfrak{S}_{F \times_H G}$ and $\mathfrak{S}_F \times_{\mathfrak{S}_H} \mathfrak{S}_G$. It suffices to verify fiberwise that $\mathfrak{S}_{F \times_H G}(S) \cong \mathfrak{S}_F(S) \times_{\mathfrak{S}_H(S)} \mathfrak{S}_G(S)$ for all $S \in \mathfrak{S}$. By lemma $3.9$, $\mathfrak{S}_{F}$ is a fibered category over $\mathfrak{S}$, hence we can use lemma $3.31$ which applies to fibered categories and get $\mathfrak{S}_{F \times_H G}(S) \cong \mathfrak{S}_F(S) \times_{\mathfrak{S}_H(S)} \mathfrak{S}_G(S)$ for all $S \in \mathfrak{S}$. We get an 1-morphism, isomorphism $\alpha: \mathfrak{S}_{F \times_H G} \cong \mathfrak{S}_F \times_{\mathfrak{S}_H} \mathfrak{S}_G$, and we denote the inverse as $\alpha^{-1}$. This is an equivalence since the 2-morphism $(\alpha^{-1} \circ \alpha) \Rightarrow Id_{\mathfrak{S}_{F \times_H G}}$ is the identity, hence a 2-isomorphism. Similarly, $(\alpha \circ \alpha^{-1}) \Rightarrow Id_{\mathfrak{S}_F \times_{\mathfrak{S}_H} \mathfrak{S}_G}$ is a 2-isomorphism.
Recall/Summary (Example 3.8 p17):
The 2-category $\mathfrak{S}_F$:
Let $F : \mathfrak{S}^{opp} \to Categories$ be a functor (i.e. a presheaf of categories). Associate to $F$ the following fibered category $\mathfrak{S}_F$ over $\mathfrak{S}$: Objects are pairs $(U,x)$ of objects $U$ in $\mathfrak{S}$ and $x \in F(U)$. Morphisms from $(U, x)$ to $(V, y)$ are pairs $(f, \varphi)$ of morphisms $f : U \to V$ and $\varphi : x \to f^* y$, where we write $f^∗ := F(f)$. The composition of $(g, \psi : y \to g^∗z) \circ (f, \varphi : x \to f^∗y)$ is defined as $(g \circ f, f^∗(\psi) \circ \varphi)$. The projection to $\mathfrak{S}$ forgets the second component of the pairs.
There seem to be two points of confusion in the question.
Point 1: Why are the only 2-morphisms of categories fibered in sets the identities?
Well, what is a 2-morphism of fibered categories?
Let $A$ be the base category, $P:B\to A$, $Q:C\to A$ fibered categories (over $A$), $F,G:P\to Q$ 1-morphisms of fibered categories (i.e. functors such that $QF=QG=P$). Then a 2-morphism $\alpha:F\to G$ is a natural transformation from $F$ to $G$ with the property that $Q(\alpha_b)=1_{Pb}$ for all $b\in B$ (i.e., $\alpha_b$ lies in the Q-fiber over $Pb$ for all $b \in B$).
In the case that $Q$ is fibered in sets, since $\alpha_b$ is always in the $Q$-fiber over $Pb$ (which is discrete/a set), we have that $\alpha_b$ is an identity morphism. Since $\alpha_b:Fb\to Gb$ is an identity morphism, we conclude that $Fb=Gb$ for all $b\in B$, and for all $f:b\to b'$, the naturality squares force $Ff=Gf$, so $F=G$, and $\alpha=1_F=1_G$.
In other words, if $Q$ has discrete fibers, then the hom categories $\mathbf{Fib/A}(P,Q)$ are also discrete.
Point 1.5: Implications of Point 1 for 2-fiber products vs 1-fiber products
Claim: if $R:D\to A$ is a fibered category with discrete fibers, and $P:B\to A,Q:C\to A$ are arbitrary fibered categories, and $F:P\to R$, $G:Q\to R$ are 1-morphisms of fibered categories, then the 1-fiber product $P\times_R^1 Q$ is in fact the 2-fiber product $P\times_R^2 Q$.
Here's a simple proof. Suppose I give you a 2-commuting square $$ \require{AMScd} \begin{CD} T @>>> P \\ @VVV @VVV \\ Q @>>> R, \\ \end{CD} $$ then because $R$ has discrete fibers, the only 2-morphism that can make this square commute is an identity, so it actually 1-commutes. Thus there is a unique morphism $T\to P\times_R^1 Q$. Uniqueness of this morphism guarantees uniqueness up to isomorphism, so this makes $P\times_R^1 Q$ satisfy the universal property of a 2-fiber product of $P$ and $Q$ over $R$.
Alternatively, just check that when $R$ has discrete fibers, the explicit construction of $P\times^2_R Q$ reduces to something isomorphic to the usual construction of $P\times^1_R Q$.
Point 2: Why does this fact imply the claimed result?
I'm going to use $\int U$ to denote the category of elements/Grothendieck construction for $U:A^{\text{op}}\to \mathbf{Cat}$, since this is more standard notation in my experience, at least for presheaves valued in sets.
We want to show $$\int U\times_{\int W} \int V\cong \int U\times_W V$$ where $U\overset{\phi}{\to} W \overset{\psi}{\leftarrow} V$ is a cospan of presheaves of categories, and $W$ is valued in discrete categories.
We know that the fiber product on the left can be taken to be the 1-fiber product when $W$ is a presheaf in $\mathbf{Set}$. Then the objects on the left hand side are tuples $((a,u),(a,v))$ with $u\in U(a)$, $v\in V(a)$, such that $\phi(u)=\psi(v)$, and morphisms from $((a,u),(a,v))$ to $((a',u'),(a',v'))$ on the left hand side are tuples $((f:a\to a',\alpha : u\to f^*u'),(f,\beta: v\to f^*v'))$, such that $\phi(\alpha)=\psi(\beta)$.
On the other hand, objects on the right hand side are tuples $(a,(u,v))$ with $(u,v)\in (U\times_W V)(A)=U(A)\times_{W(A)} V(A)$, and morphisms $(a,(u,v))\to (a',(u',v'))$ on the right hand side are pairs $(f:a\to a', (\alpha,\beta):(u,v)\to f^*(u',v'))$.
Comparing the data, we see that the two sides consist of the same data, and we can give an isomorphism between the two categories.
End note
When $U$ and $V$ are also presheaves valued in sets, this becomes even simpler, since morphisms on the left are now just $f:a\to a'$ such that $u=f^*u'$, $v=f^*v'$, and morphisms on the right are also $f:a\to a'$ such that $(u,v)=f^*(u',v')$.