Before the collision both rocks are flying parallel to the x-axis, so they have zero velocity in the y-direction (which we’ll abbreviate as vy0, where the 0 denotes your reference frame):

vy0 = 0 for both rock A and rock B

therefore total y-momentum is zero before the collision.

Symmetry tells you that in the collision, the rocks are deflected by equal and opposite amounts. Therefore as rock 1 moves by 1m in the y-direction and rock 2 moves by 1m in the opposite direction to reach their respective target lines, they take equal times to do so (t0(A) = t0(B)  = t).  So their y-velocities will then be:

v+y0 = 1/t for rock A and  - 1/t for rock B

(recall that we're using the superscript + to mean "after the collision").

Their momenta in the y-direction will therefore be: p+y0  =  m/t for rock A and -m/t  for rock B.

The total y-momentum after collision is therefore m/t - m/t  =  0:  this is the same as it was before the collision, so momentum was conserved.

Note that the y-velocities of the two stones after the collision would actually be affected by the relativistic effects of stone A hurtling away from you and stone B towards you. However this does not invalidate the result:  the stones have equal speeds relative to you (just in opposite directions), so the relativistic corrections would be the same for both and hence the total classical momentum after the collision would still be zero.