Power can't give you the pull force. It depends on how fast you are pulling something. To pull something twice as fast at the same pull force, you need twice as much power. And vice versa. To pull something with twice as much force at the same speed requires the same doubling of power.
Alternatively, but following the same logic as above, you can only get torque of an electric motor if you know at which RPM you are going to need that torque. Power divided by angular velocity will give you torque.
The current times voltage is, indeed, power consumed by the motor. You do have to subtract the power wasted to heat, however. So the full formula is IV - IR², where R is the resistance of the coil.
Finally, keep in mind that simply because the motor is rated at 2.8A and 115V, it doesn't mean it will be drawing 2.8A at 115V. The amount of current the motor draws will depend on the RPMs the motor is going at and applied voltage. The formula for current is I=(V-kω)/R, where R is aforementioned resistance, V is applied voltage, ω is angular velocity, and k is a constant unique to the motor. If you know the maximum RPM the motor reaches under no load and given voltage, you can estimate that constant by k=V/ωmax.
Replacing magnets in a motor will effectively alter the constant k. R will remain the same, unless you change the coils as well. And these two constants effectively determine performance of an electric motor under ideal conditions. Naturally, real world tends to be slightly more complicated, but it's a very good estimate to start off with.
Hope some of that helps.
