Let me ask you a question instead. The consistency of PA is known to be independent of PA so we have a model of PA that thinks there is a proof of something contradictory like $0 \neq 0$. Therefore, this "proof" that $0 \neq 0$ is in the set-theoretic universe $V$. So does $V$ think that Peano Arithmetic is inconsistent?

The answer is no because $V$ realizes that this is not a true proof but rather a proof involving nonstandard numbers either with formulas or length. The same type of idea is happening here. Even if we have a nonstandard model thinking that it has a polynomial time algorithm for SAT, a standard model looking at this algorithm may see things differently. For an even more extreme example, consider the fact that if we take a total computable function $f$ with any given running time, a nonstandard model computes the standard portion of it in a time amount that it views as constant because it has a fixed $c$ that's greater than every Natural number. But does this mean that the function can actually be computed in constant time? Of course not, because $c$ is not a true finite number.

I should also make mention that the last thing I said is of a slightly different nature since even the nonstandard model will not view itself as computing the function in constant time. Mainly, it does not know where the standard portion ends and the nonstandard part begins.

If P = NP turned out to be independent of ZFC, then we'd have a model of ZFC that would think that P = NP since by the definition of independence, P $\neq$ NP would not be provable from the axioms of ZFC. However, this would not be sufficient for generalizing the result to all models as you conjectured since there would also have to be a model of ZFC thinking that P $\neq$ NP by virtue of ZFC not proving P = NP. These results follow directly from GĂ¶del's completeness theorem. On the other hand, if P = NP were provable in ZFC, then all models of ZFC would think that P = NP.
Let me ask you a question instead. The consistency of PA is known to be independent of PA so we have a model of PA that thinks there is a proof of something contradictory like $0 \neq 0$. Therefore, this "proof" that $0 \neq 0$ is in the set-theoretic universe $V$. So does $V$ think that Peano Arithmetic is inconsistent?
The answer is no because $V$ realizes that this is not a true proof but rather a proof involving nonstandard numbers either with formulas or length. The same type of idea is happening here. Even if we have a nonstandard model thinking that it has a polynomial time algorithm for SAT, a standard model looking at this algorithm may see things differently. For an even more extreme example, consider the fact that if we take a total computable function $f$ with any given running time, a nonstandard model computes the standard portion of it in a time amount that it views as constant because it has a fixed $c$ that's greater than every Natural number. But does this mean that the function can actually be computed in constant time? Of course not, because $c$ is not a true finite number.