Assume that the top left corner is black (this causes a loss of generality, but we'll compensate by proving that both colors can be reassembled). Sum the lengths of the black parts along the top edge of the square and let the result be $x$; similarly define $y$ for the left edge. Then the area of the black parts is $xy + (2-x)(2-y) = 4 - 2x - 2y + 2xy = 2$. Rearranging, we have $2(x-1)(y-1) = 0$, so at least one of $x = 1$ or $y = 1$ holds. If $x = 1$ we can simply cut out the black (or white) rectangles and push them horizontally until they touch the next black (white) rectangle; this yields horizontal strips all of width $1$, whose heights sum to $2$. If $y = 1$ just push the rectangles together vertically instead, and rotate the resulting big rectangle by $90^\circ$.