There is no property of $T$ alone that will ensure that $T+S$ always has an $\omega$ model in the circumstances you describe. In fact, there is no computably axiomatizable theory $T$ with the property that $T+S$ has an $\omega$ model whenever it is consistent. To see this, suppose we have such a $T$, and simply let $S=\neg\text{Con}(T)$. It follows that $T+S$ is consistent, but it can have no $\omega$-model, since the proof of a contradiction that $S$ asserts must be nonstandard.

What you need is a property about $T+S$, not just a property about $T$.

There is no property of $T$ alone that will ensure that $T+S$ always has an $\omega$ model in the circumstances you describe. In fact, there is no computably axiomatizable theory $T$ with the property that $T+S$ has an $\omega$ model whenever it is consistent. To see this, suppose we have such a $T$, and simply let $S=\neg\text{Con}(T)$. It follows that $T+S$ is consistent, but it can have no $\omega$-model, since the proof of a contradiction that $S$ asserts must be nonstandard.

What you need is a property about $T+S$, not just a property about $T$.

Regarding your revised question mentioned in the comment, what we seem to need is a characterization of which theories have $\omega$-models.

Theorem. A theory has an $\omega$-model if and only if it is consistent in $\omega$-logic.

See the Wikipedia entry on $\omega$-logic. Note that being consistent in $\omega$-logic is not the same as being $\omega$-consistent. The latter has to do essentially with a single instance of the $\omega$-deduction rule, whereas consistency in $\omega$-logic allows nested applications of the rule.