If you have a 100W electrical load and you drive 100W plus efficiency losses, say 110W, into the generator, things will be in a state of equilibrium, with 100W being converted from mechanical input power into electricity, and the other 10W of mechanical input power being eaten up by losses.
Now suddenly put 1kW of mechanical power into the machine; at that instant, before the rotational speed can change, the 100W electrical load will continue to present the same mechanical load to the prime mover. Things will not be in equilibrium, and the machine's rotational speed will accelerate. Depending on circumstances, this may or may not increase the electrical load. Certainly the generated voltage will go up, and any simple resistive load will therefor absorb more power, but maybe you have some regulation such that the load continues to draw exactly 100W.
So assume the load continues to draw exactly 100W. Where does the extra 900W of mechanical power go then? The machine's speed must increase until the losses equal the mechanical driving power; so it ends up turning extremely fast, the increased power going into increases in friction in the bearings, windage loss due to the rotating parts, eddy currents in the magnetics (and doubtless a couple of other things I forget at the moment), none of which are desirable.
You would find that, without exceeding the machine's electrical rating, you would quickly exceed its mechanical ratings, i.e., probably long before you got to 1000W, the rotation speed would be several times the suggested speed, and catastrophic failure would likely result. Note you can do this with no electrical load on the generator at all.
For your second question - generally - yes. Constant speed means no acceleration. Acceleration is proportional to the torque. Acceleration zero means torque (sum of total torques) zero. As for the first one, you can use an energy equilibrium equation:
$$E_{electrical}=E_{mechanical}$$
$$E_{electrical}=V\cdot I$$
$$E_{mechanical}=\tau\cdot\omega$$
Plus applicable loses. Here V and I are voltage and the current, tau and omega are the torque and angular velocity.
(Well, strictly speaking these are not energies, but power, but since we are interested in a momentary values, the equality still valid).
Best Answer
Of course having the generator produce electrical power puts a load (torque in opposition to rotation) on the shaft. Without such load, there would be no mechanical power transferred into the generator, and it would violate conservation of energy by producing electrical power.
Electric generators, which can be thought of as mechanical to electrical power converters, can be made reasonably efficient. Good ones are 90% efficient or more. In that case, 90% of the load on the shaft is due to the electrical power produced, and the other 10% is due to losses. These losses include mechanical friction, electrical losses in the wires, and magnetic losses in the core material and due to induced currents in unwanted places.