In Katz and Mazur's book "Arithmetic Moduli of Elliptic Curves" (available here), the following statement is made about full set of sections and fibered product (Lemma 1.8.5, page 35).
Let $Z_1/S$ and $Z_2/S$ be two non-empty finite flat $S$-schemes of finite presentation and ranks $N_1$, $N_2$ respectively. Let $$P_1^{(1)},\ldots ,P_{N_1}^{(1)}\in Z_1(S)$$ $$P_1^{(2)},\ldots ,P_{N_2}^{(2)}\in Z_2(S)$$ be given sequences of $N_i$ points in $Z_i(S)$ for $i=1,2$. Suppose that $Z_1/S$ is finite étale. Then the following conditions are equivalent:
1. For $i=1,2$, the $N_i$ points $P_1^{(i)},\ldots ,P_{N_2}^{(i)}$ in $Z_i(S)$ form a full set of sections of $Z_i(S)$.
2. The $N_1N_2$ $S$-valued points $P_j^{(1)}\times P_k^{(2)}$ of $Z_1\times _S Z_2$ form of full set of sections of $Z_1\times _S Z_2$.
(For the definition of a full set of sections, please look at the book or read this related question).
After the proof of this statement, it is stated that if we drop the hypothesis that $Z_1$ is étale over $S$, then we only have $1. \Rightarrow 2.$
I have been trying to find a counterexample to disprove $2. \Rightarrow 1.$, but I have been unable to find one. I tried to look at cases where $S=\operatorname{Spec}(k)$ is the spectrum of a field, and $Z_1$, $Z_2$ the spectra of non-reduced $k$-algebras (so that they are not étale), such as $k[\epsilon]/(\epsilon^2)$, but I couldn't find anything satisfying. Sadly, my intuition about étaleness is not developped enough to have a feeling of what could work here.
Would someone here please be able to provide an example?
I thank you very much for your help.
I noticed later that in fact, two examples are given in (1.11.4) and (1.11.5), pages 52-55 in the book. I am going to write the second one down as an answer to my own question.
Let $p$ be a prime, and $S=\operatorname{Spec} (\mathbb{Z}/p^2\mathbb{Z})$. We let $Z_1=Z_2= \mu_p$ be the $p$-roots of unity group-scheme over $\mathbb{Z}/p^2\mathbb{Z}$, that is $\mu_p=\operatorname{Spec}((\mathbb{Z}/p^2\mathbb{Z})[X]/(X^p-1))$. We then define the two group morphisms $$\phi_1,\phi_2:\mathbb{Z}/p\mathbb{Z}\rightarrow \mu_p(\mathbb{Z}/p^2\mathbb{Z})$$ by $\phi_1(a)=\phi_2(b)=1$. They are the zero maps. The section $1$ is the evaluation at $X=1$ ring map. We denote by $\phi$ the induced morphism $\mathbb{Z}/p\mathbb{Z}\oplus \mathbb{Z}/p\mathbb{Z} \rightarrow \mu_p(R)\oplus \mu_p(R)$ defined by $\phi(a,b)=(1,1)$ which is again the zero map.
These group morphisms define sets of sections of $\mu_p$ and $\mu_p\times \mu_p$ simply by looking at their images. It happens that $\phi_1$ and $\phi_2$ do not define full sets of sections of $\mu _p$. Indeed, if we look at the element $f=X^p-a\in (\mathbb{Z}/p^2\mathbb{Z})[X]$, we see that the matrix of the multiplication-by-$f$ endomorphism in the free $\mathbb{Z}/p^2\mathbb{Z}$-module $(\mathbb{Z}/p^2\mathbb{Z})[X]/(X^p-1)$ of rank $p$,with basis $1,X,\ldots, X^{p-1}$, has determinant $1-a^p$, which is different from $f(1)^p=(1-a)^p$ for at least one value of $a$ in $\mathbb{Z}/p^2\mathbb{Z}$.
In other words, we do not have the following equality of Cartier divisors in $G_m$ over $\mathbb{Z}/p^2\mathbb{Z}$: $$(X-1)^p\not = (X^p-1) \quad\text{in}\quad (\mathbb{Z}/p^2\mathbb{Z})[X]$$
However, $\phi$ actually defines a full set of sections of $\mu_p \times \mu_p$ over $\mathbb{Z}/p^2\mathbb{Z}$. To show this, we need to verify the following statement:
For any algebra $R$ and for any $f(X,Y)\in R[X,Y]/(X^p-1,Y^p-1)$, we have the identity $\operatorname{Norm}(f)=f(1,1)^{p^2} \operatorname{mod} p^2R$.
For this, we may view the norm as the composite of $$R[X,Y]/(X^p-1,Y^p-1)\xrightarrow{N_X}R[Y]/(Y^p-1)\xrightarrow{N_Y}R$$
First, note that in $G_m$ over $\mathbb{Z}/p\mathbb{Z}$, we do have the equality of Cartier divisors $$(X-1)^p = (X^p-1) \quad\text{in}\quad (\mathbb{Z}/p\mathbb{Z})[X]$$
Hence, the zero morphism actually defines a full set of sections of $\mu_p$ over $\mathbb{Z}/p\mathbb{Z}$. It follows that we have the norm identity for the $\mathbb{Z}/p\mathbb{Z}$-algebra $(R/pR)[Y]/(Y^p-1)$, which results in the congruence $$N_X(f(X,Y))=f(1,Y)^p+pg(Y)$$ with some $g(Y)\in R[Y]/(Y^p-1)$.
Now, for any two elements $A,B \in R[Y]/(Y^p-1)$ and an indeterminate $T$, the norm with respect to $Y$ of $A+TB$, that is the image of $A+TB$ by the map $R[T][Y]/(Y^p-1)\xrightarrow{N_Y} R[T]$, is a polynomial in $T$ of degree $p$, of the form $$N_Y(A+TB)=N_Y(A)+\sum_{i=1}^{p-1}T^i\Lambda_i(A,B) + T^pN_Y(B)$$
On the other hand, $\operatorname{mod} p$, we have $$N_Y(A+TB)=(A(1)+TB(1))^p=A(1)^p+T^pB(1)^p \quad\operatorname{mod}p$$
Equating coefficients, we obtain $\Lambda_i(A,B)=0\quad\operatorname{mod}p$ for $i=1,\ldots ,p-1$ and therefore $N_Y(A+TB)=N_Y(A)\quad\operatorname{mod}(pT,T^p)$.
Applying this with $T=p$, we conclude $N_Y(A+pB)=N_Y(A)\quad\operatorname{mod}p^2$.
Eventually, $$\begin{align} \operatorname{Norm}(f(X,Y))&=N_y(N_X(f))\\&=N_Y(f(1,Y)^p+pg(Y))\\&=N_y(f(1,Y)^p)\quad&\operatorname{mod}p^2\\&=N_Y(f(1,Y))^p \quad&\operatorname{mod}p^2\\&= (f(1,1)^p+pk)^p \quad&\operatorname{mod}p^2\\&= f(1,1)^{p^2}\quad&\operatorname{mod}p^2 \end{align}$$