If N>2, it is well known that if two invertible NxN matrices A and B have the same determinants of any 2x2 corresponding submatrices, then A=B or A=-B.  

Given then all these 2x2 determinants of an invertible matrix A, is there an "explicit" way to recover/write down A?  

If N=3 it is easy, as you can get the determinant of A (up to the sign) and all its cofactors, so you can obtain the inverse matrix of A or -A, but when N>3?



2x2 subdeterminants of a matrix
By "corresponding submatrices" I presume you mean those $2\times2$ minors obtained by deleting $n-2$ colums and $n-2$ rows, where these columns and rows have the same $n-2$ indices. Once you've calculated the determinants of these submatrices you recover the action of $A$ on the exterior square $\Lambda^2 V$.


Now the following paper: "An algorithm for recognising the exterior square of a matrix" by Catherine Greenhill explains how to then obtain the original matrix $A$. Here's the relevant quote:



> 
> One computational problem which presents itself immediately is this: how can we
>  determine whether a given matrix $Y$ is equal to the exterior square of another matrix $X$?
>  In particular, if such an $X$ exists then we would like to construct one. A polynomial-time
>  algorithm which solves this problem is described in Section 5.
> 
> 
> 


The paper can be downloaded [here](http://web.maths.unsw.edu.au/~csg/papers/ext-matrix.pdf).


One needs to be slightly careful here, because the exterior square does not quite determine the matrix $X$ uniquely. Here is another quote from the paper:



> 
> We prove in Section 4 that two matrices $X$, $X'$ with rank at least three
>  have the same exterior square if and only $X'\in \{X, -X\}$.
> 
> 
> 


So if the rank is at least three (which it is, since you are assuming invertibility), then we are pretty much done. I'm guessing that the situation where the rank is $\leq 2$ would be easy enough to resolve but in any case that's outside the scope of the question...


