AC/DC - Current generator with short circuit
The current of Ig2 must flow through the short circuit between node 4 and 5. It has no other influence on the rest of the circuit.

To see why, just apply superimposition, leaving only Ig2 active. You'll have Ig2 with its terminal shorted, so no voltage across them. Anything else connected between node 4 and 5 cannot be powered, since it is passive and its terminals are connected to a "0V source".

You can actually remove Ig2 from your circuit, but you must add its value to I, because in the first circuit the value of I comprises some component due to the rest of the circuit (let's call it Ix) minus Ig2. The remaining part of the circuit will work the same.

The equation will be:

I = Ix - Ig2 => Ix = I + Ig2

where I assume Ix has the same direction as I. In other words, Ix is your I in the second circuit.