Now this could be naive, but I'm inclined to think that any of the shifting around as a result of changing the angle of declination for a pushup shouldn't matter in terms of the net external force the hands have to produce. I would wager any change of measurement you find on a scale has more to do with fluids shifting in the body, and thus the center of mass changing.
For example, if one considers a very large simplification of the body as a rigid body(not a bad approximation if the athlete has good core strength) with two torques (F_g = weight, F_n = normal) acting on an axis of rotation, you can see why. Suppose there are two situations, one in which the 'rigid body' is parallel with the ground (normal pushup), and the second in which the 'rigid body' is tilted with respect to the ground:
In both situations the gravitation force/weight acts through the center of mass, and the force of 'pushing' is felt via the normal force in the rigid body. In the second situation where the rod has been rotated, the torque about the axis has changed, but it changes the same for both the weight and normal vector.
This means that relatively, nothing has changed between the forces, you've made gravitational torque less 'strong' but also the normal torque less 'strong' to the same degree. Your arms will have to push with the same force as they were in the parallel case. However, this does mean there is now a compressive component to the forces, but the internal forces don't matter for a rigid body, nor would effect the torque the arms in a pushup would have to produce. The mechanics on the arms change I'm sure, and probably change the moments on the elbow and shoulder, but the overall net force required to rotate the rod/body has not changed. Therefore, I think any change in weight measure on a scale would be from a non-mechanical influence.