This seems to be an old conjecture of Shafarevich. This much is mentioned in Barry Mazur's article in "Modular Forms and Fermat's Last Theorem."

My apology for sounding argumentative in relating my impression from Mochizuki's papers: it was not my intention, and I should not have used the word "disagree."

I agree. Let us keep on examining Mochizuki's papers, and hope that his daring but beautiful strategy (deforming, as it were, number fields!) really works. ABC is such a delicate problem: the slightest error in one part of the theory could easily ruin the whole proof (as with Miyaoka's earlier attempt). As for the asymptotic discrepancy with ABC near-misses, Mochizuki himself seems to be a bit perplexed (see the remark on p. 40); hopefully this will be straightened out.

On closer look, I see that the assumptions of Thm. 1.10 are indeed a bit more complicated: for the field $F$ obtained by adjoining $\sqrt{-1}$ and the $3 \cdot 5$-torsion points of $E$, it must be that $G_F \to \mathrm{GL}_2(\mathbb{F}_{\ell})$ is surjective. Still, a stronger version of uniform open image conjecture should make this automatic, as soon as $\ell >> 0$. Anyway, I apologize if my post violates the rules of MO: if so, please feel free to remove it from this thread (or delete it).

With $F=\mathbb{Q}$ and $E$ as above we have $F_{\mathrm{mod}} = \mathbb{Q}$ and $F_{\mathrm{tpd}}=F_{\mathrm{mod}}(E[2])=Q$. Then note that the first displayed inequality before the proof on p. 23 involves only $F_{\mathrm{tpd}}$, not $F$. That the ABC-triple is primitive is the essential assumption of the ABC-conjecture. As for semistability, it certainly holds away from 2; let's say we assume $16∣ab(a+b)$ as you say: it does not make much of a difference for the shocking outcome.

Indeed, I was afraid this was probably not the place to write this, as it has nothing to do with the original question.