EDIT: This is a completely new answer.
I will prove that your specific suggestion of defining a Hilbert class field of an elliptic curve $E$ over $K$ does not work. I am referring to your proposal to take the smallest field $L$ such that the corestriction (norm) map $\operatorname{Sha}(L) \to \operatorname{Sha}(K)$ is the zero map. (I have to assume the Birch and Swinnerton-Dyer conjecture (BSD), though, for a few particular elliptic curves over $\mathbf{Q}$.)
Theorem: Assume BSD. There exists a number field $K$ and an elliptic curve $E$ over $K$ such that there is no smallest field extension $L$ of $K$ such that $\operatorname{Cores} \colon \operatorname{Sha}(L,E) \to \operatorname{Sha}(K,E)$ is the zero map.
Proof: We will use BSD data (rank and order of Sha) from Cremona's tables. Let $K=\mathbf{Q}$, and let $E$ be the elliptic curve 571A1, with Weierstrass equation
$$y^2 + y = x^3 - x^2 - 929 x - 10595.$$
Then $\operatorname{rk} E(\mathbf{Q})=0$ and $\#\operatorname{Sha}(\mathbf{Q},E)=4$. Let $L_1 = \mathbf{Q}(\sqrt{-1})$ and $L_2 =\mathbf{Q}(\sqrt{-11})$. It will suffice to show that the Tate-Shafarevich groups $\operatorname{Sha}(L_i,E)$ are trivial.
Let $E_i$ be the $L_i/\mathbf{Q}$-twist of $E$. MAGMA confirms that $E_1$ is curve 9136C1 and $E_2$ is curve 69091A1. According to Cremona's tables, $\operatorname{rk} E_i(\mathbf{Q})=2$ and $\operatorname{Sha}(\mathbf{Q},E_i)=0$, assuming BSD. Thus $\operatorname{rk} E(L_i) = 0+2=2$ and $\operatorname{Sha}(L_i,E)$ is a $2$-group. On the other hand, MAGMA shows that the $2$-Selmer group of $E_{L_i}$ is $(\mathbf{Z}/2\mathbf{Z})^2$. Thus $\operatorname{Sha}(L_i,E)[2]=0$, so $\operatorname{Sha}(L_i,E)=0$.
This solution works iff $q$ is a prime power, so it's not a full solution, but I guess it's better than nothing, as it is completely elementary.
If $q$ is a prime power, then $q$ must fully divide either $s$ or $s-1$. We may WLOG assume it divides $s$ (since otherwise we may take $t=1-s$ as $s$). Then we have that, letting $s=mq$,
$$2(q-1)^2(q+1)=m(mq-1).$$
Taking $\bmod q$, we have that $m\equiv -2\bmod q$, so we may set $m=(x+1)q-2$, giving $s=-2q+(x+1)q^2$ and
$$(x^2+2x-1)q^2 - (4x+2) q + (5-x) = 0,$$
as noted in the question. From this, we have $x\equiv 5\bmod q$. Let $x=nq+5$. Then
$$s=-2q+(nq+6)q^2=nq^3+6q^2-2q=6q^2+q(nq^2-2).$$
We will prove the following lemma:
Lemma: If $q$ is a nonzero integer and
$$(2s-1)^2 = 8(q-1)^2q(q+1)+1,$$
then $|s|<3q^2.$
Proof: First, note that
$$3x^4-(x-1)^2x(x+1)=2x^4+x^3+x^2-x=(2x^2-x)(x^2+x+1)$$
is positive for all $x$ outside of $[0,1/2]$, and specifically is positive if $x$ is a nonzero integer. Thus,
$$(x-1)^2x(x+1)<3x^4.$$
$$(2s-1)^2=8(q-1)^2q(q+1)+1<24q^4+1\leq 25q^4.$$
So,
$$|2s-1|<5q^2.$$
$$|2s-1|+|1|<5q^2+1\leq 6q^2.$$
$$2|s|< 6q^2,$$
implying the result.
If $n=0$, then $s=-2q+6q^2$, and we have
$$(6q^2-2q)(6q^2-2q-1)=2(q-1)^2q(q+1).$$
This quartic has a triple root at $q=0$ and a single root at $q=11/17$, so we can safely ignore this case. Thus, $|n|\geq 1$. If $|q|\geq 11$, then $|nq|\geq 11$. So,
$$|nq^2|\geq 11|q|.$$
$$|nq^2|\geq 9|q|+2.$$
$$|nq^2-2| \geq 9|q|.$$
But
$$s=6q^2+q(nq^2-2).$$
So, since $|nq^2-2|\geq 9|q|$, we have that $|s|\geq 3q^2,$ which contradicts our lemma. Thus, we have reduced the problem to a finite case check on integers $-10\leq q\leq 10$, which can easily be done.
Best Answer
I don't see any contradiction: the Selmer group also has a contribution of rational points. Indeed, the group of 2-torsion rational points on this elliptic curve is isomorphic to $\mathbb Z/2\mathbb Z$ and generated by $(-54,0)$, as Magma will readily confirm (probably, I used GP/Pari).
UPDATE: As David suspected, the Tate-Shafarevic group of this elliptic curve is isomorphic to $(\mathbb Z/4\mathbb Z)^2$ so 2-descent misses the elements of order 4 in Sha.