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 $p$ has a root then certainly $T$ proves $p$ has a root; and 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 $T$.