Is Adam Elga's proof for thirdism in Sleeping Beauty still considered to be sound?

What's Beauty credence for Heads when she wakes on Wednesday and doesn't remember any of her awakenings on Monday/Tuesday? If it's 1/2 what is the reason for the change from 1/3? The fact that she has forgotten the information about her awakening during the experiment?

Because of the memory erasure, I think it's best to regard Beauty on Monday, on Tuesday (if woken), and on Wednesday as different people, all "descended" from Beauty on Sunday.  But if you want to think of them as the same person, the change from 1/3 to 1/2 is a consequence of the new information from being told that it is Wednesday.  The same shift happens if she's told that it is Monday.  (If she's told it is Tuesday, she of course shifts to probability 1 for Tails.)

In the standard setup, by the way, I think it's implicit that on Monday and Tuesday awakenings, Beauty is told that it is not Wednesday (ie, that it is either Monday or Tuesday).

Suppose SB wakes up on Wednesday. She is told that it's Wednesday and then left alone to dress before she will be asked for the last time to guess Heads or Tails.  She doesn't remember any of her awakenings and her credence for Heads is once again 1/2. However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1/3. Should she expect to successfully guess Tails with 2/3 probability?

Assuming that successfully hiding a piece of paper is an unlikely event, she should think the probability of Heads is 1/3 simply because that makes a hidden piece of paper twice as likely (since then her past self will have had two chances to do this).  The message on the paper is irrelevant (assuming it doesn't contain any new arguments that she hadn't thought of on Sunday).

Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn't know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?

That's an interesting scenario.  My argument for why Beauty when not told it is Wednesday should think Tails is twice as likely as Heads is that with Tails (hence two awakenings) the probability of anyone experiencing what she is (eg, exact position of fly seen on the wall, stray thought about how good omelettes taste, ...) is twice what it is with Heads (one awakening), so standard Bayesian updating makes Tails twice as likely.  When woken, told it is Wednesday, and having memories of her last awakening, however, the probability of her experience (including memories) is the same for Heads or Tails.  The two awakenings with Tails don't double the probability of her remembered experience being whatever it is (along with  knowing it is Wednesday), since whatever she experienced on Monday is forgotten. So her probability for Heads should be 1/2.

Ape in the coat: “What's Beauty credence for Heads when she wakes on Wednesday and doesn't remember any of her awakenings on Monday/Tuesday?”

The snarky answer is “irrelevant.” The assessment of her credence, that she gives on Monday or Tuesday, is based on the information that it is either Monday or Tuesday within the waking schedule described in the experiment. She is not responsible for incorporating false information into her credence.

One thing that seems to get lost in many threads about this problem, is that probability and/or credence is a function of your knowledge about the state of the system. It is not a property of the system itself. You are trying to make it a property of the system.

Halfers are most guilty of not understanding this, since their arguments are based on there being no change to the system despite an obvious change to the knowledge (I’m ignoring how to treat that knowledge). Thirders recognize the change, but keep trying to cast the difference in terms of episiotomy when it really is much simpler.

“However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1/3. Should she expect to successfully guess Tails with 2/3 probability?”

No, since her knowledge is different now than it was then.

“Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn't know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?”

If I understand this correctly, she knows that it is Wednesday. Her credence is 1/2. She can, at the same time on Sunday or Wednesday, make an assessment of the state of her knowledge on Monday or Tuesday, and recognize how it is different. In other words, "my credence now is 1/2, but my credence then should be 1/3." The paradox you seem to be fishing for does not exist.

Two points...  

Like a number of people (including Elga), you're converting an almost-doable thought experiment into one that may be impossible in principle.  If the experiment is done with some not-yet-invented but plausible memory-erasing drug, but is otherwise realistic, Beauty will not have the same experiences when woken Monday and when woken Tuesday.  Various aspects of her sensed environment, as well as internal thoughts, will be different for the two awakenings.  We just assume that none of these differences allow her to infer the day of the week.

Second, if she should wager as if the probability of Heads is 1/3, in what sense is the probability of Heads not actually 1/3?  The only point of probabilities is to influence actions.  (Also, it is not just "some" situations where she should wager on the basis that the probability of Heads is 1/3.  That's always what she should do.)

No, much of the debate gets obscured by trying to ignore how SB's information, while it includes all of the information that will be used it separate the experiment into two observations, also is limited to the "inside information" and she is in just one of those parts. That's the purpose of the shopping spree I suggested. Correcting what you wrote, if the memory wipe is complete and SB has literally no way of knowin what information might apply to what is clearly a distinct part of the experiment, but she knows there is a different part than the current one, how can it not be "considered (by her) to be a different experiences."

For the purposes of sampling, there are two parts to the experiment. What you call them is not relevant, only that SB knows that she is in one, and only one, of two parts. In the  classic version, as Elga modifided it from the actual problem he proposed, one part must have a waking and one part has either a sleeping or a waking. Are you really claiming that the part where she is left asleep is not a part of the experiment? One that she knows is a possibility that is ruled out by her current state? And so fits the classic definition of "new evidence" that you deny? But if you really believe that, what about my version where she is taken shopping? It is even better as a classic example of evidence.

Or you could look at my answer. There, I explain how the problem you are solving actually is an "alternate formulation." And, while there are still two parts, they are equivalent so we do not need to treat them separately. The answer is unequivocally 1/3.

Or try this: Instead of a coin, roll two six-sided dice. On Monday, ask her for the probability that the resulting sum is 7. On Tuesday, if the sum is odd, ask her the same question. But if it is even, ask her for the probability that the sum is 8.

If the room has a calendar, Beauty should certainly say Pr(S=7)=1/6 on Sunday and Monday. But on Tuesday, in answer to "what is the probability of 7," she should say Pr(S=7)=1/3. But without a calendar, Pr(S=7)=1/6 can't be right. Because it might be Tuesday, and Pr(S=7)=1/3 can't be right. And it might be Monday, when Pr(S=7)=1/6 can't be right. It has to be something in between, and that something is (2/3)*(1/6)+(1/3)*(1/3)=4/18=2/9. Yes, even though this is never the answer if she knows the day. That is how conditional probability works.