Let $E/\mathbb Q$ be an elliptic curve with good reduction away from a finite set of primes in $S$. Let $\mathscr E$ be a model for $E$ over $\mathbb Z[1/S]$. Then I know two ways to prove the weak Mordell Weil theorem. One (common) version is the following:
Assume that $n$ is only divisible by primes in $S$. There is an exact sequence:
$$0 \to E(\mathbb Q)/nE(\mathbb Q) \to Sel^{(n)}(E) \to TS(E)[n]\to 0$$
where $Sel^{(n)}(E)$ is the $n$-Selmer group and $TS(E)$ is the Tate-Shafarevich group. One then shows that $Sel^{(n)}(E)$ is finite.
The other argument is to note that $\mathscr E(\mathbb Z[1/S]) = E(\mathbb Q)$ by properness assumption and since $[n]: \mathscr E \to \mathscr E$ is etale away from $S$, $\mathscr E[n] = E[n]$ is unramified away from $S$ as a Galois module and therefore, we have an injection:
$$0\to \mathscr E(\mathbb Z[1/S])/n\mathscr E(\mathbb Z[1/S]) \to H^1(\mathbb Z[1/S], E[n]).$$
By $H^1(\mathbb Z[1/S], E[n])$, I mean Galois cohomology unramified away from $S$ or equivalently, Etale cohomology for the scheme $\mathbb Z[1/S]$.
Then, we can use standard results (inflation-restriction and Hermite-Minkowski or Class field theory/Kummer theory) to show that the Galois cohomology is finite.
My question is:
What is the relation between $Sel^{(n)}(E)$ and $H^1(\mathbb Z[1/S], E[n])$? Are they equal?
No, they are not equal: $Sel^{(n)}(E)$ is smaller than $H^1(\mathbf{Z}[1/S], E[n])$. The difference between the two is that $Sel^{(n)}(E)$ consists of classes that are locally in the image of $E(\mathbf{Q}_\ell)$ at all primes, while $H^1(\mathbf{Z}[1/S], E[n])$ consists of classes that satisfy this condition at primes not in $S$ but are arbitrary in $S$.
In other words, there's an exact sequence $$0 \to Sel^{(n)}(E) \to H^1(\mathbf{Z}[1/S], E[n]) \to \bigoplus_{\ell \in S} \frac{H^1(\mathbf{Q}_\ell, E[n])}{E(\mathbf{Q}_\ell) \otimes \mathbf{Z}/n}$$ The last term is clearly finite (because $H^1(\mathbf{Q}_\ell, E[n])$ is finite for every $\ell$), so $Sel^{(n)}(E)$ is finite if and only if $H^1(\mathbf{Z}[1/S], E[n])$ is. So although these groups aren't the same, you can use either in proving weak Mordell--Weil.