EDIT: Given the change to the question, this answer is no longer relevant; I'm going to leave it up because it is worth comparing the OP to the "dual" version of the question.

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Your paragraph beginning "using the same theorem . . ." is false: it is true that for any computably enumerable theory $T\subseteq Th(\mathbb{N})$ (or $T\supseteq$Robinson if you prefer) there is a polynomial $p$ with **no** root such that $T$ does not prove "$p$ has no root." However, by contrast:

I claim that, as long as $T$ contains the ordered semiring axioms and $T\subseteq Th(\mathbb{N})$ (that is, $T$ contains only true sentences), then $D_T=D_{Th(\mathbb{N})}$.

To see this, just note that for any polynomial $p$, if $\overline{n}$ is a root of $p(\overline{x})$, then we can use the axioms of arithmetic to show that $p(\overline{n})=0$ - plugging a specific value into a specific polynomial and evaluating is not something which takes logical strength! In the other direction, since $T$ only contains true sentences, if $T$ proves $p$ has a root then $p$ has a root.

So we are done.

Note how little we had to assume about $T$.