By $\vdash P = NP$, I'm going to assume that you mean you can only use the axioms of first-order logic (i.e., formulas that define the meanings of the logical symbols such as $\forall x(x = x)$ for '=') and that "P = NP" is a shorthand for a formula in the language of arithmetic stating something nontrivial. In that case, P = NP is independent (of the axioms of first-order logic) but so is $0 = 1$. For the second formula, note that $0 = 1$ is true in a model interpreting $0$ and $1$ as the same element and false in a model interpreting $0$ and $1$ as different elements. Therefore, we can neither prove it nor disprove it from the axioms of first-order logic alone. Specifically without axioms governing the interpretations of these symbols, we cannot derive any meaningful conclusions about whether P = NP. If you assume nothing, you can prove nothing!
Therefore, you can be certain that the P versus NP problem does not ask for whether we can show $\vdash P = NP$ but instead it asks whether $T \vdash P = NP$ for some theory $T$ enforcing the obvious rules we want to hold for arithmetic such as $0 \neq 1$. Moreover, it should be noted that the problem of P = NP has already been proven to be independent of certain weak theories of arithmetic. You can see a survey of the P = NP problem in http://www.scottaaronson.com/papers/pnp.pdf. The whole point is that independence results are less valuable when we assume less, and we cannot hope to prove $P = NP$ or $P \neq NP$ from a theory adequately representing its meaning without the theory being sufficiently powerful or inconsistent.
Therefore, reasonable interpretations of the P = NP problem seem to suggest a proof from PA or ZFC. If you were to find that the problem were independent of one of these stronger theories, this would indeed be very significant. However, doing so using known techniques is essentially solving the problem itself as discussed in the following excerpt from Wikipedia:
These barriers have also led some computer scientists to suggest that the P versus NP problem may be independent of standard axiom systems like ZFC (cannot be proved or disproved within them). The interpretation of an independence result could be that either no polynomial-time algorithm exists for any NP-complete problem, and such a proof cannot be constructed in (e.g.) ZFC, or that polynomial-time algorithms for NP-complete problems may exist, but it's impossible to prove in ZFC that such algorithms are correct. However, if it can be shown, using techniques of the sort that are currently known to be applicable, that the problem cannot be decided even with much weaker assumptions extending the Peano axioms (PA) for integer arithmetic, then there would necessarily exist nearly-polynomial-time algorithms for every problem in NP. Therefore, if one believes (as most complexity theorists do) that not all problems in NP have efficient algorithms, it would follow that proofs of independence using those techniques cannot be possible. Additionally, this result implies that proving independence from PA or ZFC using currently known techniques is no easier than proving the existence of efficient algorithms for all problems in NP.