Let points X and Y be arbitrary closest point pairs on the two lines. That is, X lies on the first line, and Y lies on the second line, and the distance between those two points is the same as the distance of the two lines themselves, i.e. in the original picture, h = |X-Y|.
Let H := X-Y =: (x,y,z,0). That is, H is a direction vector between the two points and it's length |H| = h is the distance between the two lines.
Let A be the affine map in question. Since A is affine, it is representible by a matrix M, using operation A(v) = M*v.
Then the sought new distance is
h' = |A(H)| = |M*H|
If the affine map A does contain shear and therefore M is not guaranteed to be orthonormal, then one can't do much better than to compute the matrix multiplication above.
If the affine map A does not contain any shearing, the formula does simplify: For ease of notation, fix ourselves to a 3D space (although 2D or other dimensions are equivalent). We can decompose the matrix M to a form M = N*S, where S is a diagonal matrix S=diag(sx,sy,sz,1) and N consists of column vectors vx,vy,vz,t, where vx, vy and vz are normalized and t is a translation component. That is, decompose the scaling part out of matrix M to matrix S.
Since A does not contain any shearing, the matrix N is orthonormal (only rotates, and potentially mirrors and its determinant is +/-1), and the new distance is
h' = |M*H| = |N*S*H| = |S*H| = |(sx*x,sy*y,sz*z)|
