Resistance in Ohm's law
... according to the formula "I = U/R" the less resistance you have, the more current you got and accordingly the more power you drain. 


Correct.


  ... is this meaning the more load you got the less power you drain? That's a little counterintuitive, ...


The smaller the resistance the more current will flow. (That should be fairly obvious - think of the water analogy: a bigger pipe has lower resistance and will have a larger flow for a given head of water.)



Figure 1. Water "current" and flow analogy. Note that the pressure (voltage) is the same in both cases. Source: Sparkfun.

The "load" is the load on the source, the energy supply or, in the case of the water analogy, the water supply or reservoir. The lower the resistance of the load the more current will flow.



Update after question updated:


  
  An electrical load is an electrical component or portion of a circuit that consumes (active) electricpower. E.g. solenoid, motor, lightbulb etc.
  


So far, so good.


  
  The stronger load (solenoid), the more wire it has, the more resistance it has.
  


Not necessarily. It's the ampere-turns that will determine the strength of the solenoid. If you double the number of turns but keep the wire gauge and voltage the same then the resistance will double and the current will halve. Double turns x half the current gives the same ampere-turns so no improvement.

To increase the ampere-turns you could:


Increase the wire size. (Lower resistance, more current -> higher ampere-turns.)
Increase the turns with the same wire size and increase the voltage in proportion to maintain the current. (Same current, more turns -> higher ampere-turns.)
Any combination of the above.



  
  The higher resistance in circuit, the lower current flow, and the power.
  


Correct.