A third answer might seem a bit redundant, but the path I used was significantly different from the other two. There's a step in mine that eliminates a lot of the "bashy" hypothetical logic (and is, at least in my opinion, more interesting).
So, some basic deductions get you here:
And now you're presented with a question:
We have at least nine pieces accounted for already. (The 8 group may be broken up, if we happen to have all ten. Also, the top left cell of the 6 group could technically go with the 7.)
So, where do the I pieces go?
Group 5 must be an L. So 6 can't have the other I piece, or it would make an L with the 7. And none of the other groups can form an I... except for group 7. So the two I pieces are group 7, and our mysterious missing piece.
(And this means that group 8 is indeed one group as well!)
And the rest of the puzzle can be finished off with similar logic:
We've already used up both Ls (though one hasn't been fully decided yet), so the group in the upper middle must be a T.
That uses up both Ts, so the left-side group must be an S, and then the one next to it is forced to be S as well. And then the two right-hand groups must be the two Os...
The S must bend right in order to not block off an area; the mysteriously missing I piece now has a single place to go, wedged in the lower right corner; and then the top-right L is finally resolved!