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 electric potential is given by the integration of the eletric field over a path. If you think in a battery (a car battery as example) there is a field leaving the cathode and entering the anion which lose its strength for greater distances. When you connect conducting wires to both terminals, they will (almost intantenously) ionize the air around them in a way that it cancels the electric field inside. From this moment on, any integral of the eletric field that you take from any point of one wire to any point of other will lead you to the same value or, in other words, the same potential diference. Thanks to this, we say that the wires and the terminals are in the same potential.
Concerning the second part, the charge will appears all over the system indeed, we say them show off just in the plates for the convenience of the model (the ideal case). As the plates are very close to each other the field between is pretty high and as consequence more charges will gather there, making the rest of the circuit despicable.
Here is a good reference about the air ionization:
https://www.youtube.com/watch?v=DOMs7mYm_zs&ab_channel=ElectroBOOM
Best Answer
To be clear on the setup, we have an ideal battery (DC voltage source) and an ideal wire (zero resistance.
In ideal circuit theory, asking what happens when one connects an ideal wire across an ideal voltage source is essentially asking what happens when $1 = 0$.
However, if we allow that the wire has non-zero radius and non-zero length, then, when the wire is connected to the battery, there is a non-zero associated inductance $L$.
Thus, there will be a current through the circuit formed by the battery and wire and that current will change at a constant rate given by
$$\frac{di}{dt} = \frac{V_{bat}}{L}$$
Now, said inductance is likely to very small and thus, the current will rapidly become enormous so this model is of little physical relevance.
For example, if the ideal battery produces 1V and the circuit inductance is $L = 1nH$, the current 1 second after the wire is connected would be $i = 1GA$ which is roughly 1,000,000 times larger than the electric current associated with lightning.
Any physical battery has an associated short circuit current and, assuming one can manage to keep the battery intact until the stored energy is depleted, the battery will produce the short circuit current through the connected wire.
For example, the short circuit current of a typical 9V battery is roughly $4A$.