Seems to me that the confusion might be because the issue is being over-simplified. There are actually two separate issues here:
First of all, a propellor, turning at a given speed, will create a certain amount of thrust. This is the thrust which we all learn about for the aeroplane technical exam, then probably forget about immediately afterwards - it's what opposes the drag. As Mike points out, we are not talking about equilibrium here, so whatever thrust is left over after counteracting the drag is used to accelerate the aircraft. Less drag means more left-over thrust, hence faster acceleration and less time on the runway. (And, once airbourne, it means a faster rate of climb.)
The second issue is turning the horsepower into thrust. This is a combination of lots of things, the main one being the pitch of the propellor, but also the efficiency of the engine/propellor, and the forward speed of the aircraft (the faster it's going, the more windmill effect there will be on the prop, which will turn the prop/engine faster - and a faster engine generates more horsepower, which turns the prop faster....)
In his original question, Monocock states that the two props are identical, fixed pitch, and turning at the same RPM. In this case, and assuming the same Vr (which wasn't stated), the thrust will be the same in both cases. So the only thing which would explain a difference in take-off runs would be different drag, as Mike suggests.
The next question, though, is why would both props be turning at the same speed? If the props are identical, and the airspeed in both cases is identical, I would expect the prop attached to the more powerful engine to be turning faster, unless that aircraft was travelling significantly slower. In practice, a more realistic situation would be that the bigger engine would have a coarser pitch prop attached to it - and since Monocock explicitly stated that this isn't the case, if this is a real-life scenario then I can't explain it.
FFF
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