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First, notice that one can have a somewhat cheating solution, like this: You didn't say what $A$ was, but let me assume that it is consistent. There are two cases. If $\Sigma$ happens to be consistent, then let Con(Sigma) be any statement that $A$ proves, such as a tautology. If $\Sigma$ happens to be inconsistent, then let Con(Sigma) be any statement that $A$ does not prove, such as the negation of a tautology. It then follows that $\Sigma$ is consistent if and only if $A\vdash$ Con(Sigma), as desired.

Second, notice that in general there cannot be a non-cheating solution (Edit: provided $A$ is particularly simple: computably axiomatizable), since the consistency of a theory is a $\Pi^0_1$ assertion, and the provability of a sentence in an elementary theory is $\Sigma^0_1$, and there will be no way to surmount this. For example, if there is a sentence $\sigma$ such that PA proves ("PA is consistent" $\iff$ PA$\vdash\sigma$), then since PA really is consistent, it will follow that PA$\vdash\sigma$ and hence that PA proves that PA$\vdash\sigma$, since it proves all true existentials. Thus, PA will prove its own consistency, in contradiction to the incompleteness theorem.

First, notice that one can have a somewhat cheating solution, like this: You didn't say what $A$ was, but let me assume that it is consistent. There are two cases. If $\Sigma$ happens to be consistent, then let Con(Sigma) be any statement that $A$ proves, such as a tautology. If $\Sigma$ happens to be inconsistent, then let Con(Sigma) be any statement that $A$ does not prove, such as the negation of a tautology. It then follows that $\Sigma$ is consistent if and only if $A\vdash$ Con(Sigma), as desired.
Second, notice that in general there cannot be a non-cheating solution, since the consistency of a theory is a $\Pi^0_1$ assertion, and the provability of a sentence in an elementary theory is $\Sigma^0_1$, and there will be no way to surmount this. For example, if there is a sentence $\sigma$ such that PA proves ("PA is consistent" $\iff$ PA$\vdash\sigma$), then since PA really is consistent, it will follow that PA$\vdash\sigma$ and hence that PA proves that PA$\vdash\sigma$, since it proves all true existentials. Thus, PA will prove its own consistency, in contradiction to the incompleteness theorem.