 Solution

The Hilbert-Bernays-Löb derivability conditions are: if $S⊢A$ then $S⊢\square A$; $S⊢\square (A\to B)\to (\square A\to \square B)$; and $S⊢\square A\to \square \square A$. The first is an external rule about provability; the latter two are internal theorems. Standard numerical correctness and the stipulated ${\Sigma }_1$ form accompany the predicate in this question. These conditions do not include reflection $\square A\to A$.

Choose a fixed refutable sentence $\perp$ and identify ${\mathrm{Con}}_S$ with $\neg \square \perp$. Different fixed refutable choices have provably equivalent consistency sentences, because $S$ proves their equivalence to a contradiction. From $\perp \to G$, necessitation and distribution give $\square \perp \to \square G$. Since $G\to \neg \square G$, this yields $G\to {\mathrm{Con}}_S$.

For the converse, necessitate $G\to \neg \square G$ and distribute to get $\square G\to \square \neg \square G$. Positive introspection also gives $\square G\to \square \square G$. The two boxed contradictories imply $\square \perp$ by formalized propositional reasoning. Thus $\square G\to \square \perp$, and contraposition gives ${\mathrm{Con}}_S\to \neg \square G\to G$. ConsequentlyFinally the deduction theorem says that $S+G$ is inconsistent exactly when $S⊢\neg G$. The internal equivalence identifies $S+G$ and $S+{\mathrm{Con}}_S$ in their deductive consequences. Hence $S⊬\neg G$ exactly when $S+{\mathrm{Con}}_S$ is consistent. Consistency of $S$ alone need not ensure this stronger consistency condition.