The isomorphism coming from duality is correct, but it is not true that the invertible sheaf $R^1f_*O_E$ on $S$ is trivial in general.
Here is a down-to-earth construction of an example where $f_*\Omega_{E/S}$ is not free. Let $S=\mathrm{Spec}(R)$ be a domain (regular of dimension $1$ if you want) such that $2\in R^*$ and $\mathrm{Pic}(S)=H^1(S, O_S^*)$ has an element of order $4$. Represent this element by a $1$-cocycle $(u_{ij})_{ij}$ with $u_{ij}\in O_S(U_{i}\cap U_{j})^*$ where $\{ U_{i} \}_i$ is some open affine covering of $S$. By hypothesis of $4$-torsion, there exist $v_i\in O_S(U_i)^*$ such that $u_{ij}^4=v_iv_j^{-1}$ for all $i,j$.
Let $E_i/U_i$ be the elliptic curve defined by the affine equation
$$ y_i^2=x_i^3+ v_i x_i.$$
Over $U_{i}\cap U_{j}$, the elliptic curves $E_i, E_j$ are isomorphic by the change of variables
$$y_j=u_{ij}^3y_i, \quad x_j=u_{ij}^2x_i.$$
The cocycle condition on the $\{u_{ij}\}$'s insures that the $E_i$'s glue together and gives rise to an elliptic curve $E/S$.
Now let's compute the differential sheaf on $E$. On $E_i$, a canonical basis of $\Omega_{E_i/U_i}$ is $\omega_i:=dx_i/y_i$ and we have $\omega_j=u_{ij}^{-1}\omega_i$ above $U_i\cap U_j$. If $\omega_{E/S}$ is free and generated by $\omega$, then writing $\omega_i=u_i\omega|_{E_i}$ with $u_i\in O(U_i)^*$, we get $u_{ij}=u_iu_j^{-1}$. Hence the class of $(u_{ij})$ in $H^1(S, O_S^*)$ is trivial, contradiction.
One can show that over a Dedekind domain (or any ring I think), $f_*\Omega_{E/S}$ is free if and only $E/S$ has a global smooth Weierstrass equation. It is well known (?) that this doesn't hold in general over number fields.
$\newcommand{\Spec}{\mathrm{Spec}}$
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
$f:E\rightarrow S$ is proper and flat. By Cohomology and Base Change, $R^1f_*(I([0])^{-n})= 0$ implies that $f_*I([0])^{-n}$ is locally free. You can check on a fibre to see that the rank is $n$.