It is important that the positions have to be legal.
Think about what might have been the last move.
Think about how the pieces got to the squares they occupy.

a1: white king. he'll be in check on all the given squares but this one seems the most sensible. b4: black king, more sensible of the two unchecked squares...so then I guess b5 black bishop because...well I'm more or less completely guessing now, can't make sense of anything. I look forward to some more constructive answers!

Ra6 threatening a8#, if rook is taken, qa8#, if king moves, qf7#.

The real question is if black can castle after 1.Ra6 because then there is no mate.

Pawn structure for black does suggest that the last move can't have been e3, (that pawn had to travel along the black squares from the a file) which means that the white king can't be on f2.

answer: W:Ka1, Bb5 B:Kf2, Nb4

B has captured at least 4 pieces on dark squares b6-e3. The only White dark-squared pieces that are missing are R,R,Q,N.

Therefore White's KR was captured on one of those 4 squares. In order to get there, at some point White must have played f2xg3, R escapes, g2xf3 (or the reverse). White has also captured a piece with his a-pawn which is on b3.

So White has captured at least 3 pieces of the following: R, N, N, Q. Black has at most one of those pieces left, in addition to his king, making at most 2 Black pieces missing. Since there are 4 pieces unaccounted for, and White has at most 2 (K and KB), both sides have exactly 2.

White's light-squared bishop can only stand on b5.

On b4 or f2 the White king would be in illegal check: Black could not have put his bishop to c3 by any legal last move, and we know the e3-pawn came from d4 last, and not e4 or f4. Therefore the White king is on a1.

Therefore White is in check. Black could not have moved his rook last (he couldn't capture on a5, because the only possible piece not accounted for by the e3-pawn is the light-squared bishop, and there is a piece on b5 so he couldn't have come from the fifth rank), therefore White is in discovered check.

Black could not have given discovered check by moving away a rook or a queen. There has always been a White pawn on b2, so the Black king never went to a3. Last move, there was a White bishop on b5 and a White pawn on b3, so the Black king could not have been on a4 (illegal check). Therefore Black did not give discovered check with Ka3-b4 or Ka4-b4. So he did it with a knight, which now stands on b4. The Black king must stand on f2.

The pawn structures tell everything! But the players could have been idiots, you can't rely on them playing good moves or standard openings. All you should assume is that they made legal moves.

Black has captured exactly 2 pieces with his c4-pawn: these must be the two bishops. Since it captured on both a light and a dark square, the pawn went a7xb6-b5xc4. White has captured at least one piece, on a3, and it was a dark-squared piece AND it was before the White QB got out to be captured by Black's pawn. It could only have been a R. However it was captured when the Black pawn was still on a7. It could only have got there via h8 and h6. Therefore Black has moved both rooks and cannot castle, so White can mate by 1. Ra6.

white king can only be on a7 (It's pretty hard for it to get there, but possible). On c7 there's double check, on g7 the bishop can't get to h8 on h7 the knight can't get to f8.

The only way the white king can be on a7 is if a black knight moved from a6 to c7. So blacks missing pawn has been promoted to a knight.

Whites missing pawn must have been promoted to a rook. Otherwise it's impossible for the white knight to get to a1 (because the only way for the king rook to get out is after axb. So white promoted his a-pawn and black promoted his h-pawn (by taking the white rook on g1 after white has already moved his knight to h1. The white king bishop could have been captured before or after this, which allows the white king to escape.

So pieces left for white: just one knight. But the black pawn on b6 couldn't be there without taking something, so there's actually no white pieces left.

Pieces left for black: One king, one rook and a light squared bishop. Only one of these pieces could have been taken and it had to happen on g3. So the light squared bishop is still in play and must be on h7 and the black king is on g7

Is it allowed for the two black knights to switch place?

Black's last move was Ra1xa2+

The solution is complete.

It can't be Ra1xa2 because there couldn't have been a white piece on a2; they are all acounted for. Light-squared bishop was taken on f1, knight was taken on b6, kingside rook was taken on g1, and the a-pawn was promoted to another rook.

No. They are on b8 and g8 and haven't moved the entire game.

Black has very few moves initially. In fact, the third move has to bee hxg6 so that Black has some legal moves.
More hints:

There is no way that a black knight could escape h1, so this cannot be true. Also, the rook could have taken a white piece (probably a white knight) by moving Ra1xa2.

W: Ka2, Bc7, Qh8. B: Ra8 last move: Rxe1

Black has only moved his pawns and KR. His first three moves were b7-b6, c7-c6 in some order, then hxg6 capturing White's Q or KN. All of his subsequent moves were made with the KR. He must have made at least one move with the KR, so it is not on h8.

White's king could not have got to a8, h8 or c7, so it is on a2.

No White knight or bishop could have got to h8, so the piece there is the White Queen, and White's KN was captured on move 3.

No White knight or bishop, or Black's KR, could have got to a8 (because Nc7 would be check) so the piece there is Black's QR.

The black R cannot be on c7, so it is White's N or B. So the Black R is not on the board. However it must have been captured with White's last move, and it is now Black's move, as otherwise Black would have no legal last move. So the N is not on c7, because it did not capture a rook there. So it is the White bishop.

White's last move was to capture the Black rook. Not with the queen on h8, since the rook did not return there, not with the rook on a1 which could not return there, not with a pawn or the bishop on c7. Not with the bishop, since it could have come only from f1, but it has left f1 before for the KR to escape. Not with the king, since whether the king was on b1 or b2 Black would still have had no legal previous move. So he captured it with the R on e1.

White king is on a2 obviously.

Remaining pieces: two black rooks, two white knights, one white queen and one white darksquared bishop.

Black has captured a white piece on g6. It takes a white piece at least 3 moves to get there, which means that black must have played b6 and c6 as the first two moves.

With one white piece captured, there are still too many white pieces left, so the black king rook must have moved and then have been captured. The only piece that can go to h8 is the white queen. This also means that the piece captured on g6 was a white knight.

Remaining: Black queen rook, white knight and white bishop.

Bishop can't go to a8, but neither can the knight, since the only way to get there is via c7 (which would force qxc7). So the black queen rook is still in it's place.

Meh, that's all I got. Can't figure out if it's a knight or a bishop on c7 (I suppose it's the knight. How else would we know what the last move is?) so I'm overlooking something or I messed up somewhere.

on c3, black's last move can only have been b4xc3 e.P. and this black pawn has to disappear. White's last move was Kb3xc3.