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).
Recall that, for node voltage analysis, a floating voltage source (a voltage source that does not connect to the GND node) poses a problem since you cannot write an equation relating the current through to the voltage across.
What you must do then is enclose the floating voltage source in a supernode, which reduces the number of KCL equations by one, and add the equation relating the voltage difference between the nodes the voltage source is connected to.
Now, the dual of node voltage analysis is mesh current analysis and here we have the dual problem when we have a current source common to two meshes - we can't write an equation relating the current through to the voltage across a current source.
What must be done then is to form a supermesh which reduces the number of KVL equations by one and add the equation relating the difference of the mesh currents to the common current source.
So, write KVL counter-clockwise around the supermesh consisting of the two voltage sources and the two resistors
$$V_1 = I_aR_1 + V_2 + (I_b - I_c)R_2$$
You have, by inspection (no KVL required for this mesh - this is dual to no KCL required for the node connected to a non-floating voltage source)
$$I_c = -1.25A $$
You need one more equation which is the equation relating to difference of the two mesh currents with the common current source.
$$3A = I_a - I_b $$
Now, you have 3 independent equations and 3 unknowns.
Best Answer
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.