If two signals have different distributions that you know and can model, can you use that to separate them?
If two signals have different distributions that you know and can model, can you use that to separate them?

Statistically, yes. Does that mean you can be certain at any point in time? No. But it allows you to transport data.

For example, if I have a mixed signal composed of signal A added to signal B, and I know that the histogram of signal A is non-Gaussian in the time domain, and that the histogram of signal B is Gaussian in the time domain, is there a way to isolate either A or B using this information?

Not to separate them, but to use both to transport data independetly: For example, you can switch the Gaussian source on and off, and detect the presence of it.

I understand that I can use MLE or some similar method to fit a Gaussian to signal B

Uff, you don't &quot;fit&quot; a Gaussian to something that has a distribution like a Gaussian. I know this sounds like nitpicking, but in these contexts, it's very important to make a distinction between the shape of the signal and the signal's amplitude distribution.

My initial idea was to create a Gaussian filter that matches the distribution of signal B, and use that to separate B from A+B, but I'm not sure how I would implement that in practice, or if that even makes sense.

Doesn't make sense, far as I can tell.
I'm afraid you'll have to look into information theory a bit, and there into what mutual information is, how we model that, and noise.
You can then make general statements on the ability of A to transport data in the presence of B, considering the latter &quot;noise&quot;. It also works the other way around!
This directly leads to the successive cancellation decoder, where you first (try to) decide the better receivable variable, then subtract the decision from the observation, then decode the other. For example:
Your A is discretely uniformly distributed on ${-2, -1.9, -1.8, \ldots , -0.1, 0, +0.1,\ldots +2\$, your B is Gaussian with zero mean and variance 0.01. You observe  Y = A+B = -1.52; you decide that by far the most likely value for A is -1.5, and with that knowledge you can infer that B must have been -0.02.
You needed distribution knowledge of A and B here to decide which of the two to decode first, and then to make a maximum-likelihood decision on A.
This successive method gives you a lower bound for your ability to get data from A+B given their distributions. However, that's not the same as a joint decoding. General, for a maximum likelihood estimator, you'd need to know the joint distribution of A and B under A+B; then you can, for any A+B, tell the most likely A and B. There's to my knowledge no general formula for this decoding – it really depends on the distributions. Also note that if A and B are statistically independent, this joint density will fall apart into the product of the individual densities. Note, however, also that usually, both A and B will employ channel coding, so that these likelihoods are both time-variant and depend on other decisions – welcome to the wonderful world of coding theory.