The answer is simply because observed massless particles always obey $E = pc$
It helps to first see what is going on here: A and B are boost/rotation transformations which leave invariant the momentum vector which represents a massless particle propagating in the z direction.
Now, why don't they change the momentum?
Well, the A and B, acting on a momentum in the z-direction, both represent a centrifugal acceleration K (orthogonal to the z-momentum) compensated by a counter rotation J. The total effect of A (as well as B) is therefor simply zero.
One can actually always compensate an orthogonal boost operation in this way with a rotation in the other direction, also for particles with mass, so
this group here is actually not uniquely for mass less particles only.
What links it to massless particles is the ratio between the K and the J and the fact that this ratio works for any momentum in the z direction. This brings us to the answer: The J and the K nullify each other because of the fixed ratio between E and p, a property of massless particles.
The general case is. $A\cos \theta ~+~B\sin\theta$
Where $\theta$ is the angle which determines the direction in the x-y plane of the centrifugal acceleration. Small nitpick about his signs: In a right-handed
coordinate system they should be:
$J_3 ~,~~~~ A ~:=~ J_2 - K_1 ~,~~~~ B ~:= -J_1 - K_2$
Note that if you reverse the direction of the momentum, (k,0,0-k) instead
of (k,0,0,k), that the signs of the generators also change. In this case you
$J_3 ~,~~~~ A ~:=~ J_2 + K_1 ~,~~~~ B ~:= -J_1 + K_2$
One should expect this because under spatial inversion K behaves like a
vector and J like a pseudo vector.