Assume that we have weakly solved chess, in the sense that we have an algorithm that when used, allows us to force a win if possible, and if that is not possible, force a draw. Define a ply as optimal as one made in accordance with such an algorithm, and further define a game as optimal if every ply made by both players is optimal.
How many such games are there?
TL;DR
The answer is that it depends on whether we solve chess for White, Black, or a Draw. If one of the first two, probably very, very many. If instead chess is drawn, then we don't know, and it's not clear how we would go about figuring this out without actually solving chess in the first place.
Pedantic setup
First, the general strategy is the same as the one pursued by Shannon's number. Estimate the number of optimal plies available in a given position, O, and estimate the number of unforced plies in an optimal game of chess, p. The number of optimal games of chess is given by the expression O^p.
Define a winning ply as a ply that either (i) wins the game, or (ii) leads to a board state in which, no matter their opponent's reply, a winning ply exists. Similarly define a drawing ply. Finally, a losing ply is defined as a ply that, if made, leads to a board state in which her opponent has at least one winning ply. Note that because chess lasts a finite number of moves, these options are exhaustive. If all plies from a position are losing, then the player to move is in zugzwang.
Weakly solving chess entails being able to show that there exists either a winning or drawing ply from the starting position, or conversely that White is in zugzwang. Solving chess entails knowing that some particular ply is winning or drawing, or some particular reply from Black is winning.
Finally, note the the definition of optimal says nothing about what to do if a position is losing. The losing player is free to do whatever they wish, because within the context of optimal play, all plies are equally bad.
If White is winning
Suppose that we solve chess for White. Since White has access to a winning ply, all of Black's 20 legal replies are losing. Thus, there are at least 20 games of optimal chess. This argument applies recursively to each resulting board state, and the number of optimal games grows quite quickly. Even if White has access to only one winning ply on each of her turns, Black always has access to all legal plies from the position. We could stipulate that optimal play in a losing position is to resign, in which case there is either 1 optimal game of chess — the winning ply from the starting position — or 0, because no one ever agrees to a game. For comparison, think about whether you would agree to a game of Tic-tac-toe.
All that remains is to estimate the length of an optimal game. Strictly speaking, knowing this number requires solving chess in the first place. However, it seems quite unlikely to me that the average optimal game is short. Arguments to this effect are difficult, however. As pointed out by @wimi in the comments, if Chess is solved for White than the very short game — 1. e4 e5 2. Qh5 Ke2 2. Qe4# — is an optimal game.
For the purposes of this answer, suppose that an optimal game lasts 30 moves. Derive a lower bound by assuming that White only ever has one winning ply. Effectively, her turns don't contribute number of optimal games, since she only every has one option. As argued before, Black has access to all legal plies in any position. The number of legal plies in a position is estimated to be about 30. Thus, a conservative lower bound on the number of optimal games is 30^(30) = 2x10^44. Different estimates of the length of an optimal game will lead to correspondingly different answers.
It gets a tricky if we place further restrictions on optimal play. For instance, you could stipulate that if one is in a losing position, the optimal ply is the one that extends the game the longest. This raises the exponent of the estimate, since games last longer, but it also knocks down the number of optimal moves available in a losing position, since they won't all extend the game the same amount. The interaction between these two is unpredictable.
If White is losing
Exactly the same situation as before, but the positions reversed. This time, no matter which of the 20 opening moves she makes, White is guaranteed to lose.
If White is drawing
Suppose instead that chess is drawn. Based on the rate of draws seen at high level computer play, I deem this the most likely. This means that there exists at least one ply for White for which there exists at least one drawing reply, and all other plies are losing. How many drawing plies are there in the opening position? Strictly speaking, we don't know. We don't even know if there is one, and figuring this out would be a major advancement in our knowledge of the game.
For instance, it could be the case that there is only one drawing ply from the starting position, and all others are losing. Further, this could be true of all resulting positions, in which case there is 1 optimal game of chess. I think this is unlikely, but strictly speaking it is possible.
Instead, it could be the case that all of White's opening plies are drawing, and all of Black's replies are as well. Then there are probably many games of optimal chess. Of course, this has to stop being the case at some point, as evidenced by the existence of both the Scholar's Mate and the Fool's Mate, so a more precise estimate is difficult.
Based on Zermelo's theorem:
There is at least one optimal game.
Let's see why. Zermelo's theorem applies to finite two-person games of perfect information in which the players move alternately and there is no influence of chance in choosing your moves. Chess is a game of this type:
Then, for all such games, Zermelo showed that with optimal play one of these results is true:
Since one and only one of these conditions is true, there is at least one optimal game.
PS: This article on Zermelo's theorem may also be of interest.
