Does a polynomial solution for an NP-complete problem that can only be implemented for small N *still* imply P=NP?
If you mean that the polynomial-time algorithm only works for inputs up to some fixed size, it shows nothing at all. Any problem at all (even if it's undecidable, let alone NP-complete) becomes a finite language when restricted to instances of constant size. All finite languages can be decided in constant time.

If you mean that the polynomial-time algorithm works for all inputs but it's still so inefficient that current computers can only run it on small instances then, sure, that would prove that P$\,=\,$NP. The idea that "polynomial" means "efficiently and effectively solvable" is just a short-hand. The actual definitions of P and&nbsp;NP don't depend on that short-hand.