Would any current flow when batteries and resistors are set up like this?
OK, since the OP says it's not homework...

As @Olin has pointed out, you have three voltage sources, each in series with a resistor, that are all connected in parallel.

The easiest way to analyze this is via Thevenin and Norton equivalent circuits. The (voltage source + series resistance) subcircuits which are Thevenin equivalents (presumably to be a first-order approximation to a battery), can be transformed to Norton equivalents:

3V + 2 ohm series -> 1.5A || 2 ohm, since 3V/2ohm = 1.5A

2V + 1 ohm series -> 2A   || 1 ohm

4V + 3 ohm series -> 4/3A || 3 ohm

Then, the Norton equivalents can be paralleled easily by adding the currents and computing the parallel equivalent resistance: 1.5A + 2A + 4/3A = 29/6A; 2 ohm || 1 ohm || 3 ohm = 1/(1/2+1+1/3) ohm = 1/ (11/6) ohm = 6/11 ohm.

To get the open circuit voltage, compute the Norton current * Norton resistance = 29/6A * 6/11V = 29/11V = 2.636V

To get the current flowing through the 1 ohm resistor and 2V source, just compute (29/11V - 2V) / 1 ohm = 7/11A (= 0.636A)