How do I calculate the current in parallel branches?
You cannot directly apply E=IR to this because the LED is a diode, which is a nonlinear device.

Simplified: a diode will not conduct current unless sufficient, correct-polarity voltage is present across its terminals to forward-bias it. 

The resistance of the switch shorting the diode out is very small, so the voltage generated across it is also very small, certainly many orders of magnitude too small to forward bias the diode.

If you replace the switch with a resistor, things can change. Imagine the LED is out of the circuit. If the resistor limits the current sufficiently to develop a voltage drop across it that is equal to or greater than the forward-bias requirement of the LED, once you put the LED in the circuit you will see that the LED will be on dimly as you observed. There is current being 'shared' by the LED and the resistor- you will observe that the voltage across the resistor in parallel with the diode is 'clamped' by the diode.

Diodes are not intrinsically resistive like resistors are. Their resistance is extremely small - this is why an LED circuit requires a series resistor - to provide resistance which limits the current and protects the diode from failing.

See the Wikipedia article on diodes.