Well, at least to the right-hand branch of circuit. In the loop formed in the left-hand branch by closing the switch, there is still 1V pushing current around in a loop through the single resistor, but it has no effect on the rest of the circuit.
Another way of thinking about the problem is that when you close the switch, you make the two points indicated by black dots at the same potential. Therefore, anything contained inside those two points has no effect on what's outside.