The 30% and 70% winning percentages can only come when multiples of ten games have been played, so there must be at least 20 games. But if there WERE 20 games, they’d either have to be 3/10 and 14/20 or 7/10 and 6/20. In the first case there would need to be 11 wins in 10 games (impossible) and in the second we’d need -1 wins in 10 games (also impossible) so it can’t be done with fewer than 30 games.

It’s worth noting that all solutions must come in pairs – if every W is replaced by an L and vice versa, then every percentage P becomes (100-P) and so the 30 and 70 swap and the 40 and 60 swap, and the 50 remains unchanged. So we can assume that the 30% occurs at the 10 game mark WLOG.

If 30% is 3/10, we can earn 50% and 40% along the way if there’s one win in the first two games, a second win in the next 3 games and the third win in the following 5 games. Then 50% will occur after game 2, 40% after game 5, and 30% after game 10. We’ve further got 70% by having 21/30, so we need only to place 60%. That might come as 9/15, 12/20, or 15/25. But if we’ve only won 3/10 we can’t win enough games to get to 9/15. Similarly, if we’d only won 15 by game 25, we can’t win enough to reach 21 by game 30, so that’s out. 12/20 works, though, if we win 9 from games 11-20, and then win an additional 9 from games 21-30. 

So 30 appears to be the minimum number of games. Let’s introduce the notation {W,4*L} to mean, “one W and 4 L in any order”. Then one solution is:

{W, L}{W, 2*L}{W, 4*L}{9*W, L}{9*W, L} which gives percentages after each {} of 50%, 40%, 30%, 60%, 70%

And its pair is:

{L, W}{L, 2*W}{L, 4*W}{9*L, W}{9* L,W} which gives percentages after each {} of 50%, 60%, 70%, 40%, 30%

NOW THE EXTENSION

Suppose we add 10, 20, 80, and 90% milestones. Taking the one of the pair of solutions that starts low as before, then 10% = 1/10, and we can pick up 50% and 20% on the way if the win is in the first two games so we see 1/2 and 1/5. 30% then becomes 6/20. If we have 5 wins in games 11-20, we can reach 6/15 = 40% along the way, so after game 20 we’ve seen 10, 20, 30, 40, and 50%. 

Now we have to win most of the rest of the games. We’ve already lost 14 games, so we have to win 8 more games to get back to 50% (happening in game 28 at 14/28.) 60% at 30 is out of reach, but 21/35 is possible if we win the next 7 games. Not enough room to get to 28/40 after that for 70% and the next opportunity is 35/50 which requires 14 wins in the next 15 games. Now, we’ve lost 15 games so far, so for that to be 20%, we’d need to be at 75 games total, so we must win the next 25 games to reach 60/75 and 80%. Finally, those 15 losses are 10% when there are 150 games total, so winning the next 75 games takes us to 135/150 and the desired 90%. 

Using the same notation, this solution is {W,L}{3*L}{5*L}{5*W}{5*L}{15*W}{14*W, L}{25*W}{75*W} with win percentages after each {} of 50%, 20%, 10%, 40%, 30%, 60%, 70%, 80%, 90%. Again, swapping W and L in the solution gives its solution pair, whose win percentages P are each (100-P) of the original.

I realize I haven’t proved that 150 games is minimal, but I’m fairly confident that it's correct. Maybe somebody else can prove it?