From my understanding of the problem, you're making it harder on yourself by thinking about a physical manifestation...
Among the abstractions we make in circuit analysis are ideal, independent sources. In the schematic you've posted, the alternator is considered an ideal, 35A current source. This means that whatever it's connected to, it will deliver 35A no matter what. As such, a current source that is not part of a closed loop is a contradiction (because no current can flow through an open circuit).
Similarly, that 12V voltage source is an ideal voltage source, that will always have 12V across its terminals. Similar to a current source without a loop between its terminals, a voltage source with its terminals shorted is a contradiction (as anything connected between two ends of a wire in a schematic are supposed to be at the exact same potential).
With that said, it's not particularly "correct" to think of the electrons coming out of the current source to have a specific potential energy. You lose a lot of the physical world when you abstract a circuit to a simple schematic like posted.
In the schematic posted, and assuming the sources ideal, given that the load has 30A passing through it, then there are 5A passing through the battery (the 12V source and the 100m\$\Omega\$ internal resistance) since KCL must be satisfied. Given that the power delivered/absorbed by a device is defined to be the product of voltage and current, it should be clear that for a given current, different voltages will result in different powers. In this case, the total battery voltage is 12V + (100m\$\Omega\$)(5A) = 12.5V. Thus the battery is, in this case, absorbing positive power with a value of (12.5V)\$*\$(5A) = 62.5W.
With a 6V battery, assuming the same conditions (meaning the load is still pulling 30A and the internal resistance of the battery is identical to the 12V battery--which means the load must be different, by the way), you can see how the power absorbed by the battery would be smaller.
For further thinking... consider a simple loop consisting of a voltage source and a current source, any value for either (for instance, say 1V and 1A, respectively), and both of them connected following the passive sign convention (current source pointing from (-) to (+) on voltage source). By the definitions put into place in circuit analysis, there must be 1A flowing through the loop, so the 1V source is absorbing 1W of power, while the 1A source is delivering 1A of power. Change the 1V source to, say, a 5kV source, and the power absorbed/delivered will be greater by a factor of 5000. Note that the electrons coming out of the current source now must have 5000 times the potential (by definition, since the voltage source has 5000 times the potential across it), but the current source hasn't changed. Naturally, if the 1A current source were not ideal, it may not be able to supply 1A through a 5kV battery, but in circuit analysis, these ideal sources are defined to deliver their rated spec, regardless of the circuit they are a part of (short of contradictions).
Nope. You're missing the fact that a battery can only supply a certain amount of current - that is, it has an internal resistance. Then your circuit looks like
simulate this circuit – Schematic created using CircuitLab
So there is an internal node which you cannot see (it's not actually a single point - it's part of the overall functioning of the battery) for which KVL applies.
Please note that, if possible, you should not try too hard to apply batteries to your thinking about circuits. They are very, very non-ideal devices. A voltage/resistor model does not do the device justice. The internal voltage depends on how fully charged or discharged the device is, and the resistance depends on things like state of charge, current level and temperature.
With that said, you should realize that, for instance, a 9-volt battery has a pretty high internal resistance - it won't provide much current. A car battery, on the other hand, has a very low resistance, and if you mess around with one you can do things like weld large wires together.
Best Answer
This would not break Kirchoff's law, but it would break the LED. Take a look at the current-voltage curve for a typical LED:
Here, around the forward operating voltage (Vd) the current (i) increases vastly with tiny changes in voltage. Vd in your case is 2V, typical for standard red, green and yellow LEDs. At 20V (not shown on the graph) the theoretical current would be phenomenal! This would be enough to vaporise most of the LED's constituent elements. In reality, what would happen is that the LED would shine very brightly for a very short time indeed before letting out its "magic smoke". This is the very reason that we include current limiting resistors in LED circuits.
If you swap the 12V battery for a 2V battery, things get interesting. Now we're right where the dotted line above Vd sits - small changes in voltage give large changes in current. Here, the current is defined by an exponential equation with a strong dependency on temperature. We won't know what the actual operating current is, but it will be strongly dependent on the exact battery voltage and the temperature, along with the battery's non-ideal series resistance (acting a bit like the series resistor we normally use with a diode). However no engineer worth their salt would ever drive a LED like this. LEDs are very sensitive to voltage changes, and their brightness (the parameter we want to control) is strongly related to their current. Hence LEDs are always driven with current in mind - be that with a statically chosen resistor or a specialised circuit with a feedback loop designed to keep current constant.