I am not completely sure that I understand what it is you are looking for, but let me give an answer in the hope that you find it helpful.
Remember that base change is constructed (e.g. in Hartshorne) first in the case of morphisms of affine schemes, and the general case then follows by a rather involved glueing process. So to avoid the complications, I will modify your example to the case of affine schemes instead, by taking the open subscheme $\{ Z \neq 0 \}$ inside your family of curves.
Let me also change the names of your fields so that $k$ is the field we start with, and $K=k(a,b)$.
We then have a family $$ X \subset \mathbf A^4_k$$ cut out by the equation $$ax^2+by=0;$$
in other words $$X = \operatorname{Spec} \left( \frac{k[x,y,a,b]}{I} \right) $$
where $I$ is the ideal generated by $ax^2+by$.
Now the projection morphism $$\pi : X \rightarrow \mathbf A^2 \\ (x,y,a,b) \mapsto (a,b)$$
corresponds to the map of rings
$$ k[a,b] \rightarrow \frac{k[x,y,a,b]}{I}.$$
On the other hand, the inclusion of the generic point
$$ \iota: \operatorname{Spec} K \rightarrow \mathbf A^2_k$$
corresponds to the ring map
$$ k[a,b] \hookrightarrow K$$
embedding the polynomial ring in its field of fractions.
Finally, the crux of the matter: the base change of $\pi$ along $\iota$ is (by definition) given by the tensor product
$$ K \otimes_{k[a,b]} \frac{k[x,y,a,b]}{I}$$
which is isomorphic to
$$\frac{K[x,y]}{J}$$
where now $J$ is the ideal in $K[x,y]$ generated by the polynomial $ax^2+by$.
So the generic fibre is indeed defined by the same equation, but now viewed as a polynomial over a different field.
I hope that helps.
$\newcommand{\Spec}{\mathrm{Spec}}$
Scholze---> Katz-Mazur: I really wouldn't stress too much about this, to be honest. Probably Scholze should say that $p$ is locally of finite presentation and/or $S$ is locally Noetherian. Since the moduli spaces of such objects constructed is locally Noetherian, you really have no harm restricting to such a thing. Then, proper implies finite type and since S is locally Noetherian this implies that $p$ is locally of finite presentation. And then, yes, we use
[Tag01V8][1] If it makes you feel any better, his ultimate goal with this paper, and subsequent ones (which, incidentally, my thesis is a generalization of one of these papers) is to work in the same realm as the work of Harris-Taylor. In Harris-Taylor's seminal book/paper where they prove local Langlands for $\mathrm{GL}_n(F)$ they explicitly restrict only the schemes which are locally Noetherian (as does Kottwitz, if I recall correctly, in his original paper "On the points of some Shimura varieties over finite fields).
Katz-Mazur ---> Scholze: A smooth proper connected curve over a field is automatically projective. We may assume we're over $\overline{k}$. Let $X$ be a smooth proper conneced curve. Let $U$ be an affine open subscheme. Then, by taking a projectivization of $U$ (i.e. locally closed immerse $U$ into some $\mathbb{P}^n$ and take closure) and normalizations you can find an $X'$ which is smooth and projective containing $U$. Then, you get a birational map $X\dashrightarrow X'$. One can then use the valuative criterion to deduce this is an isomorphism.
An elliptic curve is connected. Note then that if $X/k$ is finite type, connected, and $X(k)\ne \varnothing$ then $X$ is automatically geometrically connected. Since any idempotents in $\mathcal{O}(X_{\overline{k}})$ must show up at some finite extension, it suffices to show that $X_L$ is connected for every finite extension $L/k$. Note that since $\Spec(L)\to \Spec(k)$ is flat and finite then same is true for $X_L\to X$, and thus $X_L\to X$ is clopen. Thus, if $C$ is a connected component of $X_L$ it's clopen (since $X_L$ is Noetherian) and thus its image under $X_L\to X$ is clopen, and thus all of $X$. Suppose that there exists another connected component $C'$ of $X_L$. Then, by what we just said the image of $C$ and $C'$ both contain any $x\in X(k)$. Note though that if $\pi:X_L\to X$ is our projection, then $\pi^{-1}(x)$ can be identified set theoretically as $\Spec(L\otimes_k k)=\Spec(L)$ and co consists of one point. This means that $C$ and $C'$, since they both hit $x$, have an intersection point. This is a contradiction. So an elliptic curve, being connected and having $E(k)\ne \varnothing$, is automatically geometrically connected.
Best Answer
Yes. Here is an argument provided by Kay Rülling, I hope that this is helpful for someone else:
It is sufficient (and necessary because $S$ has just two points: the special point $s$ and the generic point) to show that $f$ is smooth. By this lemma (or EGA IV, 4, 17.5.1), we use local finite presentation and smoothness of the special fiber to conclude that $f$ is smooth on an open $U \subset X$ containing the special fiber $f^{-1}(s)$.
Since $f$ is proper, we conclude that $U = X$, so in particular the generic fiber is smooth. Indeed, $X \setminus U$ is closed; if it were non empty, then the properness of $f$ implies that $f(X \setminus U) \subset S$ is closed and non-empty, and since $S$ is a discrete valuation ring, $f(X \setminus U)$ has to be equal to $s$ or $S$. In both cases $s \in f(X \setminus U)$, which implies that $X \setminus U$ meets $f^{-1}(s)$, hence a contradiction.