Consider the cyclotomic extension $k=\mathbb{Q}_p(\zeta_p)$ of the p-adic field $\mathbb{Q}_p$ and let $ \mathbb{Z}_p[\zeta_p]$ be the ring of integer of $\mathbb{Q}_p(\zeta_p)$. For the uniformizer $ \pi=\sqrt[p-1]{-p}$, the set $ \ \pi \mathbb{Z}_p[\zeta_p]$ is a maximal ideal of the ring of integer $ \ \mathbb{Z}_p[\zeta_p]$.
Then show that $$S=\{x \in \mathbb{Z}_p[\zeta_p]: x =\pi^2 x', x' \equiv -\frac{s^2}{2} \mod \pi \mathbb{Z}_p[\zeta_p]), s \in (\mathbb{Z}_p[\zeta_p])^{*} \} \subset (\pi \mathbb{Z}_p[\zeta_p])^2.$$
My approach:
We know that $(\mathbb{Z}_p)^*$ can be expressed as $(\mathbb{Z}_p)^{*} \cong \mu_{p-1} \times (1+p \mathbb{Z}_p)$, where $\mu_{p-1}$ is the cyclic group of $(p-1)^{th}$ root of unity in $\mathbb{Z}_p$ and $(\mathbb{Z}_p)^{*}=\mathbb{Z}_p-p \mathbb{Z}_p$.
In the same way $ (\mathbb{Z}_p[\zeta_p])^{*}=(\mathbb{Z}_p[\zeta_p])-(p\mathbb{Z}_p[\zeta_p])$.
Let $x \in S$, then $x=\pi^2 x', \ x' \equiv -\frac{s^2}{2} (\mod \pi \mathbb{Z}_p[\zeta_p]), \ s \in (\mathbb{Z}_p[\zeta_p])^{*}$.
But $x=\pi^2 x' \in (\pi \mathbb{Z}_p[\zeta_p])^2$ because $x'=-\frac{s^2}{2} \in (\mathbb{Z}_p[\zeta_p])^{*}$.
So $x \in S \Rightarrow x \in ( \pi \mathbb{Z}_p[\zeta_p])^2$, which implies $S \subset (\pi \mathbb{Z}_p[\zeta_p])^2$.
Can someone check my calculation ?
Does the reverse relation $(\pi \mathbb{Z}_p[\zeta_p])^2 \subset S$ hold ?
I assume that $p \neq 2$ (for $p=2$, I am not sure if the definition of $S$ even makes sense).
I claim your sets are equal for $p \equiv 1,3$ mod $8$ (i.e. $p=3, 11, 17, 19, 41, 43, ...$), but otherwise are disjoint.
Let $K$ be any finite extension of $\Bbb Q_p$, with ring of integers $A$ and uniformiser $\pi$. Then it is a standard fact (but needs $p \neq 2$) that every element of the principal units $1+\pi A$ is a square. Further, say $A/\pi \simeq \Bbb F_q$ (in your case, $q=p$), and let $\xi$ denote a primitive $q-1$-th root of unity in $K$. Then we have
$(\pi A^\ast)^2 = \pi^2 \cdot \langle\xi^2\rangle \cdot (1+\pi A)$
whereas in your case $K= \Bbb Q_p(\zeta_p)$, the set $S$ is
$S = \pi^2 \cdot (-1/2) \cdot\langle\xi^2\rangle \cdot (1+\pi A)$.
These two sets are equal if and only if $-1/2$ is a square in $\Bbb F_p^\times$; and if it's not, they are disjoint. (In particular, your line
is wrong resp. makes no sense (what is $(\pi \Bbb Z_p[\zeta])^*$?); you would need $x' \in (\Bbb Z_p[\zeta])^2$ which, because all principal units are squares and $s^2$ is a square, is equivalent to $-1/2$ being a square.)
Finally, playing around with quadratic reciprocity, one sees that $-1/2$ is a square in $\Bbb F_p^\times$ iff $p \equiv 1$ or $p \equiv 3$ mod $8$: The first case is the one in which both $-1$ and $2$ are squares mod $p$, the second case is the only one in which both $-1$ and $2$ are non-squares mod $p$, so their quotient must be a square.