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.
The answer to your question is "yes".
The relevant relationship is $I= nqAv_d$ where $I$ is the current, $q$ is the charge on a single charge carrier $n$ is the number of charge carriers per unit volume and $v_d$ is the magnitude of drift velocity.
In a series circuit the current is constant as a consequence of the law of conservation of charge.
This means that the amount of charge per second entering a conductor at one end must equal the amount of charge per second leaving the conductor at the other end.
Charge is not created or destroyed within the conductor.
To simplify matter let’s look at the variables one at a time and assume that the current is the same in a complete circuit or part of a circuit.
Keeping the area and the charge on a charge carrier the same means that if a conductor has fewer charge carriers per unit volume the drift speed must be larger than the drift speed in a conductor with more charge carrier per unit volume.
So how is this increased drift speed achieved?
It is achieved by having a larger voltage (per unit length of the conductor) across the conductor which has the fewer charge carriers per unit volume.
So if you had a piece of copper and piece of iron (a worse electrical conductor that copper) which had the same dimensions as the copper in series and you passed a current though them the voltage across the iron would be larger than the voltage across the copper.
Put another way the resistance of the sample of iron is larger than that of copper.
This is your resistor and copper wire situation.
Suppose now there were two copper wires connected in series of the same length but one piece of wire had twice the cross sectional area of the other.
To transport the same amount of charge per second through both wires the drift velocity in the thinner wire must be twice that of the thicker wire.
You could say that there were in total fewer charge carrier in the thinner wire than in the thicker wire and so to convey the same charge per second the charge carriers would have to move twice as fast.
How is this extra speed achieved, again by having a larger voltage across the thinner wire that the thicker wire.
The resistance of the thinner wire is greater than that of the thicker wire.
You can do a similar analysis comparing the charge on the charge carriers if they are not the same.
All this you can convert to the analogy of a flow of water with the pressure difference being the analogy of voltage and the length and cross-sectional area of pipes being analogous to the length and the cross-sectional area of the conductor.
Usually it is the volume flow per second which is cited as the analogue of current but that means that there is a problem with finding an equivalent to the number of charge carries per unit volume in the electrical case.
So it is better to say that the mass transported per second is the analogue of current.
How this is done in practice I do not know. Perhaps it has to be shown as two separate circuits?
Anyway is you have two pipes of the same dimensions and need to transport the same amount of mass per second through the pipes then the liquid with the smaller density (number of charge carriers per unit volume) would have to travel faster (drift velocity) and so there has to be larger pressure difference (voltage) across the pipe with the lower density liquid flowing through it.
An finally as a caveat to my initial answer to your question it is theoretically possible by juggling the area of the specimens (resistor and wire) to make the charge carriers in a resistor move slower than that in a wire but in practice this is not possible because the charge carrier density in copper is so much greater than that of the substances used in resistors.
Best Answer
The heat generated is the wattage dissipated, namely $W = VI$, so if the resistance is lower, the current will be higher, and if the voltage remains the same, you get more heat.
If you short out the battery (provide a very low-value resistor) then the other resistance in the circuit will take the heat, namely the battery itself, and the voltage across the short circuit will be zero. This is a good way to burn out a battery.
EDIT: OK, you're saying it is counter-intuitive that more resistance means less heat. Let me try to explain it. First, let's assume the voltage source has very low internal resistance compared to the resistor you are experimenting with, like, say, a 12-volt car battery.
Now, you put your resistor R of, say, 1 ohm across the poles of the car battery. So 12 amps will flow, so 12*12 watts of heat come off. 144 watts, that's a lot of heat. (In fact, you should probably be using big light bulbs instead of little resistors.)
Now, you take two resistors R and tie them in parallel. How much resistance do they have? R/2, because 12 amps will flow through each one, for 24 amps total. (R = V/I = 12/24 = 1/2 ohm) Each resistor sees 12 volts and 12 amps for 144 watts, but what does the pair of them see? 144 watts in each one, so 288 watts together. So by cutting the resistance in half, you doubled the heat (assuming the source has low internal resistance).