= Solution
An <arithmetic> theory is <one-consistent> if there is no <bounded arithmetic formula> $\delta(x)$ for which it proves $\exists x\,\delta(x)$ and also proves $\neg\delta(\overline n)$ for every standard <natural number> $n$. A finite tuple of existential witnesses can be coded by one <natural number>. The associated <Sigma-1 soundness> property says that every $\Sigma_1$ sentence it proves is true in the standard <natural numbers>.
Suppose first that the theory is one-consistent, but proves a false $\Sigma_1$ sentence $\exists x\,\delta(x)$. For every standard $n$, $\neg\delta(\overline n)$ is then a true bounded sentence. Bounded completeness makes each such sentence a theorem, violating one-consistency.
Conversely suppose the theory is $\Sigma_1$-sound. If it proved both the existential sentence and every negated standard instance, soundness would supply an actual witness $n$. Its positive instance $\delta(\overline n)$ is a true bounded sentence and hence provable. Together with the assumed negative instance this would make the theory inconsistent; an inconsistent theory proves a false $\Sigma_1$ sentence, contrary to soundness. Thus
$$
\boxed{\text{one-consistency}\ \Longleftrightarrow\ \Sigma_1\text{-soundness}}
$$
under the stated bounded-completeness assumption. In particular either condition implies that the theory is a <consistent first-order theory>.
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