OK, lets continue then.
A pump plus load is a coupled system.
Does the pressure generate the flow, or does the flow generate the pressure ? The point is that you cannot ask it like that.
The pump has a operating curve pressure=f(flow), you can invert the curve so that flow=g(pressure) if you like.
The load also has a similar curve.
Pressure and flow settles where the two curves intersect.
A positive displacement pump will have a curve that is very "steep" when compared against a centrifugal pump for example.
Some loads are just resistive. If you add a pressure regulator that opens at a certain pressure, then it changes the load curve.
If you block the outlet of the pump, then you have exchanged the load curve for another curve which is actually a straight line where flow=0 at any pressure. This "curve" will then intersect with the pump curve probably at a higher pressure than at the normal load curves intersection point.
Now go the other way. Completely open the outlet of the pump. How much pressure does it generate?
That would be a curve that is almost flat, but not completely flat. There will be a minute pressure.
All of the above assumes a static system. A dynamic system is more complicated.
When I said that a pump generates a pressure and not a flow, is because at a blocked outlet the flow will be zero. The pump will not generate an unlimited pressure to overcome the 'resistance'. This is a simplification of course. To explain in detail why this simplification is not correct is a tangent if you ask me.
To gas' hydraulic jack: It is a special application. Here the load is changing dynamically (as long as the piston is free to move, then the volume in the system expands when the pressure increases).
To the original topic:
If you stop the pump, and there is no accumulator, then the pressure will drop to zero "instantly" in any practical meaning of the word.
One way to save energy is to use an adjustable pump. That means that the pumps curve is variable in stead of fixed.