Let $A \rightarrow B$ be an etale ring morphism.
We have that the map, via base changing by $-\otimes_AB$ $$ B \rightarrow B \otimes_A B$$ It is etale, hence smooth hence flat.
Is this map faithfully flat?
My thoughts are applying 10.38.16, Stacks Project. It suffices to show that the induced map on Spec is surjective. I could briefly recall that
A) $B\otimes_AB \simeq B \times B'$ from etale property with multiplication $B\otimes_AB \rightarrow B$ coinciding with projection. ( I don't know why)
Combined with.
B) Given two rings $R_1, R_2$, if $P$ is prime ideal in $R_1$, then $P\times R_2$ is prime ideal in product. Indeed, if $(a,b)(c,d)=(ac,bd) \in P\times R_2$, $a$ or $c$ is in $P$.
We deduce that the map is faithfully flat.
Let $X = \operatorname{Spec} A$ and $Y = \operatorname{Spec} B$. The map you are interested in is the projection $p_1 : Y \times_X Y \to Y$ onto the first factor. As you point out, it suffices to check that this map on $\operatorname{Spec}$ is surjective.
Indeed, we see it is surjective, by observing that $p_1 \Delta = \mathrm{id}_Y$, where $\Delta : Y \to Y\times_X Y$ is the diagonal, induced from the multiplication map $B\otimes_A B \to B$.