The OP is asking about the probability of winning given that Monty has opened the door that neither player chose. Dilip Sarwate's answer describes the "standard rules" of the two-player game in which this cannot happen: Monty will always throw one player out of the game rather than open the unchosen door.
Why stick to the "standard rules" instead of answering the question? We can easily imagine a game with non-standard rules that will allow for the OP's supposition. For example:
When one player guesses correctly, Monty flips a coin (with bias $b$) to decide whether to open the unchosen door, or throw out the other player. If neither player guesses correctly, Monty flips another coin (fair, we hope!) to decide who to throw out.
In the "standard rules" game, the bias $b$ is set to 1, so Monty will always throw out a player (and you're better off staying put). We could call the other edge case "Monty's a big old softy": if bias $b$ is set to 0, Monty will never throw out a player if he can avoid it. (If he does, you know you'd better switch!)
But we don't need to know $b$ in order to solve the stated problem, because we already supposed that Monty didn't throw out a player. In that case we know that one player must have guessed correctly. If we assume that each player had equal likelihood of guessing correctly, then there is no benefit to switching, exactly as Randy argued in his answer.
We could create other non-standard rules, but the symmetry between Players 1 and 2 remains. The only way to break it is to give one player a greater chance of choosing correctly, or alternatively, allow Monty's choice of whether to throw someone out to depend on who guesses correctly. Both of these options seem unfair.
Note: I'm assuming throughout that the contestants must choose different doors.
Exercise for the reader: find the value of $b$ for which there is no benefit in either switching or staying put, even after Monty throws out the other player.