I think what is meant here is indeed not a monic polynomial over $\mathbb Z$, but rather one whose coefficients are (jointly) coprime. This, of course, is uniquely determined by the minimal polynomial of an algebraic number $x$.

Now, if $x$ satisfies the equation $$ a_nx^n + \dots + a_1 x + a_0 = 0 $$ with $a_i\in\mathbb Z$ jointly coprime, and is an $S$-unit, then for $v\not\in S$ we have $\lVert{x}\rVert_v\leqslant 1$. Say, we have $\lVert{x}\rVert_v\leqslant R$ for $v\in S$; then, analogously to your argument in the post, we get $\lVert{a_k/a_n}\rVert_p \leqslant CR^{2n}$ for all $p$-adic valuations on $\mathbb Q$. But this together with joint coprimality means that the denomitator $a_n$ is bounded: the inequality $$ \lVert{a_n}\rVert_p\geqslant \frac{\lVert{a_k}\rVert_p}{CR^{2n}},\quad k=0,\dots,n-1 $$ together with the fact that $a_0,\dots,a_n$ cannot all be divisible by $p$ implies $$ \lVert{a_n}\rVert_p\geqslant \frac{1}{CR^{2n}}, $$ so $a_n$ is indeed bounded as long as we have a bound on norms of $x$.

I think what is meant here is indeed not a monic polynomial over $\mathbb Z$, but rather one whose coefficients are (jointly) coprime. This, of course, is uniquely determined by the minimal polynomial of an algebraic number $x$.

Now, if $x$ satisfies the equation $$ a_nx^n + \dots + a_1 x + a_0 = 0 $$ with $a_i\in\mathbb Z$ jointly coprime, and is an $S$-unit, then for $v\not\in S$ we have $\lVert{x}\rVert_v\leqslant 1$. Say, we have $\lVert{x}\rVert_v\leqslant R$ for $v\in S$; then, analogously to your argument in the post, we get $\lVert{a_k/a_n}\rVert_p \leqslant CR^{2n}$ for all $p$-adic valuations on $\mathbb Q$. But this together with joint coprimality means that the denominator $a_n$ is bounded: the inequality $$ \lVert{a_n}\rVert_p\geqslant \frac{\lVert{a_k}\rVert_p}{CR^{2n}},\quad k=0,\dots,n-1 $$ together with the fact that $a_0,\dots,a_n$ cannot all be divisible by $p$ implies $$ \lVert{a_n}\rVert_p\geqslant \frac{1}{CR^{2n}}, $$ so $a_n$ is indeed bounded as long as we have a bound on norms of $x$.