Assuming $\text{PA}$ is consistent. Then $\text{PA} + \neg\text{Con}(\text{PA})$ is consistent and have a model, say $M$. We know $M$ must be nonstandard, in which case, there is a nonstandard proof of $0=1$ from ($M$'s version of) $\text{PA}$ in $M$. The fact that $M$'s version of $\text{PA}$ is different from the "real" $\text{PA}$ make the assertion $$ M\vDash\text{PA} + \neg\text{Con}(\text{PA}) $$ kind of less stunning than we might expect it to be.

My question is if we can compose some other expressions of $\text{PA}$ that if $M\vDash\neg\text{Con}(\text{PA})$ then there is a (nonstandard) proof of absurdity making use of only the standard part of $\text{PA}$ and logic axioms. If it is impossible, how to prove it?

Assuming $\text{PA}$ is consistent. Then $\text{PA} + \neg\text{Con}(\text{PA})$ is consistent and have a model, say $M$. We know $M$ must be nonstandard, in which case, there is a nonstandard proof of $0=1$ from ($M$'s version of) $\text{PA}$ in $M$. The fact that $M$'s version of $\text{PA}$ is different from the "real" $\text{PA}$ make the assertion $$ M\vDash\text{PA} + \neg\text{Con}(\text{PA}) $$ kind of less stunning than we might expect it to be.

My question is if we can compose some other expressions of $\text{PA}$ so that if $M\vDash\neg\text{Con}(\text{PA})$ then there is a (nonstandard) proof of absurdity making use of only the standard part of $\text{PA}$ and logic axioms. If it is impossible, how to prove it?