You prove that vertex cover reduces to some problem A, does that mean that A is NP-Complete?
Let's start with the relevant definitions:


  A problem $A$ is NP-complete if:
  
  
  $A$ is in NP, and
  $A$ is NP-hard.
  
  
  A problem $A$ is NP-hard if for every problem $B$ in NP, there is a polytime reduction from $B$ to $A$.


Using these definitions, you can prove the following result:


  If a problem $B$ is NP-hard and there is a reduction from $B$ to $A$, then problem $A$ is also NP-hard.


Applying this to your problem, if there is a reduction from vertex cover to $A$, then $A$ is NP-hard. If $A$ is also in NP, then it is furthermore NP-complete. Whether P=NP or not has absolutely no bearing on this.

Note that there are NP-hard problems which are provably not in NP. For example, the halting problem is NP-hard, but is known not to be in NP.