$\newcommand{\Spec}{\mathrm{Spec}}$
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.
Without loss of generality we may assume that the base scheme $S$ is Noetherian. Let $X/S$ be an abelian scheme and let $[n]:X \to X$ be the multiplication by $n$ map. Since $X/S$ is smooth it is in particular flat, and a morphism between flat $S$-schemes is flat if and only if its fibers are flat (fibral criterion of flatness). But if $s \in S$ then $[n]:X_s \to X_s$ is just the multiplication by $[n]$ map on an abelian variety over a field, which is flat. Moreover, properness of $X$ implies properness of $[n]$ and once again we can check that $[n]$ is quasi-finite on fibers, and then proper+quasi finite implies finite.
To conclude, $[n]$ is a finite flat morphism and so the kernel is a finite flat group schemes over $S$. Surjectivity can be checked fiberwise as well, so $[n]$ is an isogeny.
Best Answer
for any group variety over a field, smoothness is equivalent to being geometrically reduced.it is true because you can check smoothness after a base change to algebraic closure, over an algebraically closed field smooth locus of a reduced variety is non-empty open $U$(there are several ways to see this) and then because you have a group scheme you can translate $U$ by elements of $A(k)$ and again this would be an open smooth subvariety and obviously union of this translations is the whole space.
also always we have Smooth+connected implies geometrically integral: smooth obviously implies reduced and if you take a point on the intersection of two irreducible components the variety can't be smooth at that point(because for example, the local ring at that point could not be regular).
I'm not sure why do you think separated condition has anything to do with other hypothesis because it is needed in both definitions.