Input/Output Current from OP amp

But assuming an ideal op-amp (drawing no current) and an ideal voltmeter (drawing no current), where does the current flow to (in order to satisfy Kirchhoff's current law)?

The inputs of an ideal op-amp draw no current.

But the output behaves like an ideal voltage source --- it can source or sink as much current as needed to respond to signals at the inputs. For purposes of Kirchoff's current law you can imagine the other side of this voltage source being connected to ground inside the op-amp symbol.

Kirchoff's current law is satisfied by the output pin sinking current so that the R2 current equals the R1 current.

Welcome to the world of real opamps. Real opamps are not ideal. For this particular case - the output does not swing exactly to rails. Event for opamps that are advertised as "rail to rail" output will be several mV from the real rail.

Output swing is usually specified in the datasheet. It's called output swing, min/max output or something like that, look for output specifications. For 741 it's +-16V when powered from +-20V or +-13V when powered from +-15V. It varies with load and many different factors.

Usually when designing opamp circuits you try to opearate in the middle between the rails, and avoid going near the rail (except for comparator configurations). Note that in amlifier case opamp will start to behave funky when you swing close to a rail - you wil start getting distortion. YMMV, depends on how good your opamp is.

Another thing - you are doing inverting configuration wrong. You have to tie noninverting input to a midpoint between the rails. 2.5V in this particular case. It does not necessarily have to be the exact mid point - it all depends on your input signal.

Think about it intuitivelly - in your case for the opamp to swing high, your input signal has to go below the ground, so you set up "virtual ground" somewhere between rails so that your opamp has some reference agains which to invert the input signal.