This was quite a fun twist on weighing puzzles. The fact we don't know what the scale outputs mean requires us to, as part of our weighing procedure, decipher the scale's output. Luckily, we only need identify the fake coin and not whether it's lighter or heavier - I'm pretty sure distinguishing between lighter and heavier is impossible with only the given apparatus.
Let's label the coins as ABCDEFGHIJKL, and the three possible outputs of the scale as X, Y, Z. When weighing one set of coins against another, the order is important - the first set always goes on the left side of the scale (or whatever passes for "left").
Weigh ABCDEF against GHIJKL. This will always be imbalanced, which is a good thing in this case. Let's label the scale's output as X - we now know that X corresponds to an imbalanced state. This is the only single weighing where we know whether the scales balance or not - otherwise, there are some scenarios where we included the fake coin and some where we omitted it, so we would gain no immediate information about the scale's outputs.
Weigh AB against GH. There are only two outcomes possible - the scales remain imbalanced in the same way as weighing 1 (case a), or they now balance (case b). In case a, the scale outputs X again; in case b, we see a different output, which we will label as Y. An output of Y thus means the scale is balanced.
Case a - there are 4 candidates for the fake coin (A, B, G, H).
Case b - the fake coin is one of the other 8, but we now know when the scales are balanced, which is a huge advantage as we can now distinguish between all three possible weighing results (as any output we have not yet seen must by elimination be Z, which means the scale is imbalanced in the opposite way to X).
Case a - Weigh A against G. Again, there are only two possible outcomes. If the scale outputs X, the scales are still imbalanced, so either A or G is fake. Otherwise, A must balance with G, so either B or H is fake. Either way, there only remain two possible fakes.
Case b - This now resolves in the same way as the standard 12-coin problem, except we don't know the difference between lighter and heavier (but this doesn't matter as we merely need to identify the fake coin, we don't need to deduce if it is light or heavy). Weigh CDJ against EFI. There are three sub-cases here:
Subcase i - we get X, so the fake must be either C, D, or I, as those coins stayed on the same side as in weighing 1.
Subcase ii - we get Y, so the fake must be one of the two coins we didn't weigh here - K or L.
Subcase iii - we get Z, so the fake must be either E, F, or J, as those coins are on the opposite side to where they were in weighing 1.
Case a - we only have two candidates left (either A and G, or B and H). We can thus weigh one candidate against a known normal coin, keeping the candidate on the same side as it has been so far. If we get X - that coin is the fake; if we don't, by elimination the last candidate is the fake.
Case bi - weigh C against D. If we get X, C is the fake. If we get Y, I is the fake. If we get Z, D is the fake.
Case bii - weigh K against L. If we get X, L is the fake. If we get Z, K is the fake. We cannot get Y as one of these two is definitely fake.
Case biii - weigh E against F. If we get X, E is the fake. If we get Y, J is the fake. If we get Z, F is the fake.