Your original text admitted three interpretations, and I'm leaving the answers here:
1: What happens with a toy model when there's a circuit with an ideal battery and no resistance?
All the charge moves around the circuit at one moment in time (infinite current). The energy must leave the system as Electromagnetic radiation - accelerating charges radiate, and while that radiation would happen at any bends in the wire, it would probably happen most at the terminals where the charge goes from stopped to moving (and visa versa). The effect of energy leaving a system via EMR is often ignored in circuits, but basically, we've made a single signal broadcast antenna.
A battery represents two separate reservoirs of charge, so if charge moved between them on the terminal side, the battery would be a rock. Therefore, the work done by the current must happen between the terminals on the wire side. Don't worry too much about this scenario - there's no such thing as ideal batteries. Their internal resistance is orders of magnitude smaller than a circuit's resistance, and orders of magnitude larger than that of the wire.
2: What happens when I connect the terminals of a real battery with an ideal (superconducting?) wire?
Real batteries have internal resistance, so the battery will heat up and quickly either run out of charge or burn/explode.
3: What happens when I connect the terminals of an ideal battery with a real wire?
We specify battery voltage in circuits, so the current running through the wire will be high enough that when you multiply it by the resistance, you get back the battery's voltage. Also, the wire will heat up quickly and radiate light and heat until the ideal battery runs out of charge.
Some of the other answers included other ideas.
First, any circuit is a loop, so it will have an inductance. Inductance slows down the current in a circuit, but does not effect the circuit in steady state (or provide a real (pun intended) voltage drop). In the ideal battery/wire case, the inductance would cause the current to grow over time - that's nonsense because infinite current can't grow (you also need non-zero resistance to find the time constant - I don't divide by zero).
Second, we sometimes think of batteries as chemical capacitors. An ideal battery is not a capacitor. But if it were and there were an inductance in the circuit, the charge would move from one side to the other with an angular frequency of $(LC)^{-1/2}$. In response to the edit question, 'Will it discharge like a capacitor?', the time constant for an RC circuit is $RC$, so zero in this case. The battery won't send charges back the other direction in the circuit though because ideal batteries are not capacitors.
Incidentally, a capacitor also slows down the movement of charge, but it reverses the polarity, so the phase moves in the opposite direction. Also of note, whereas an inductor slows down changes in current most when they change most, a capacitor allows the freest flow of current when it is uncharged.
Lastly, it seems pretty clear that you're talking about a closed circuit, but if you weren't, well, nothing happens on open ideal circuits (unless they were recently closed, or will be closed soon).
Responding to other edits:
"If not, please tell me why the electrical circuit theory is meaningless without resistance."
I'm not sure what you're asking, but can we build circuits with just transistors, inductors, capacitors and diodes? I guess, but it'd be a lot more difficult to keep the magic smoke in. Circuits would also be a lot more difficult because we often model speakers, lights, motors and almost every useful thing in a circuit as a resistance. LC circuits (which have no resistors, but non-zero resistance) have a few important applications, but even so, we often put resistors in to dampen non-frequency signals or manage the voltage (with a voltage divider for example).
"If possible, can anyone give me [a fluid flow] analogy with a circuit with zero resistance, internal and external?"
I refer you to the Waterfall, though I had hoped for an aquatic image shaped more like Ascending and Descending. Water does not have an easy analogy for electromagnetic radiation because they are different phenomena. In another direction, flow of fluids is tremendously resistive, so perhaps the analogy you're looking for is that as fluids (and circuits) get colder, resistance goes down. The behaviors of both of these systems are subject to laws that are very foreign to our understanding as warm intuitioned creatures, and they won't help you in your circuits class.
Why is this energy $U$ absorbed by the battery? Why shouldn't it be
liberated as thermal energy (heat) as it happens during charging of a
capacitor?
The answer is that your analysis has no resistance in the circuit.
Assume that the initial charge on the capacitor is $Q$, the final charge on the capacitor is $\dfrac Q2$ and the emf of the battery is $V$.
You have worked out the work done in sending charge through the battery during the separation process is $\dfrac Q2 V = U$ and there is a balanced energy equation.
If there is a resistive component in the circuit then suppose that during the separation of the plates phase the potential difference across the resistive part is $v$.
This means that the total potential difference across the battery and the resistive part of the circuit is $V+v$.
As the plates of the capacitor are being separated $V+v$ must also be the potential difference across the plates of the capacitor.
This means that the force between the plates of the capacitor, which depends on the potential difference across the plates, is increased which in turns means more external work need to be done in separating the plates.
That extra work done by the external force in separating the plates is the source of the heat dissipated by the resistive part of the circuit.
Best Answer
The key word here is immediately. Since there is a $300\mathrm{k\Omega}$ series resistor between the capacitor and the battery, the series current is limited and thus, some time must elapse for the system to reach steady state.
Since the separation between the plates is quadrupled, the capacitance is reduced by a factor of 4 so, eventually, the capacitor discharges through the resistor to a charge 1/4 of the initial charge.
But in the first case, the system (effectively) reaches steady state immediately since there is no series resistance. This is best seen by adding a series resistance and taking the limit as the resistance goes to zero. You will then see that, in the limit, there is an infinitesimally brief, infinitely large discharge current that immediately brings the system into steady state.
Of course, for a physical battery, there is non-zero internal resistance and, further, the plate separation of a physical capacitor cannot be change instantaneously.