Godel's perhaps-unfortunately-named completeness theorem gives a precise positive answer: if $T$ is a set of (first-order) sentences and every model of $T$ satisfies the (first-order) sentence $\varphi$, then in fact there is a proof of $\varphi$ from $T$. More snappily, we have $$T\vdash\varphi\iff T\models\varphi$$ (technically completeness is the right-to-left direction, with the left-to-right direction being soundness, but soundness is so trivial it's often subsumed by completeness).
In particular, let $T$ be "any abstract system of axioms" (as long as they're first-order) and let $\varphi$ be $\perp$, the always-false sentence. Then "$T\models\varphi$" means "Every model of $T$ satisfies the always-false sentence," which is another way of saying "$T$ is unsatisfiable;" meanwhile, "$T\vdash\varphi$" is just another way of saying "$T$ is consistent." Contrapositing, we get that if $T$ is consistent then $T$ is satisfiable (= has a model).
(Note that "consistent iff satisfiable" is equivalent to the completeness theorem as stated above, since we can shift from "$T\not\vdash\varphi$" to "$T\cup\{\neg\varphi\}$ is consistent" and similarly for $\models$. However, this trick doesn't work in logics without negation, so in general there is a real difference here. But this is a side issue.)
There are two crucial caveats here:
It is absolutely essential that we stick to first-order sentences. In general, logics admitting a completeness theorem are quite rare.
It is also absolutely general that we look at all models of $T$ in the definition of $T\models\varphi$. The common phrasing of Godel's incompleteness theorem as "There are true statements of arithmetic which aren't provable" may appear to contradict the completeness theorem, but of course it doesn't: the issue is that "true" here refers to truth in one particular model, and that's not something that the completeness theorem has bearing on.
