I've solved the puzzle, but there seem to be two very slightly different possibilities for the final answer, differing only in what happens with two L-tetrominoes at the top in the middle:
Edit: now that the question has been slightly edited to add another grey square, there's only one possibility left, which is the second of the two above. As @Stiv notes, this means the grey squares
spell the word ODE.
The first and easiest things to fill in are the
I-tetrominoes, because there are only two columns and four rows which contain I cells. Then in the 4th row, all cells between the two I-tetrominoes must be empty (grey). In the bottom three rows, there's a T and an L between the two I-tetrominoes. The T can't go into the 4th column, and the L must be below the T in the 3rd column, so we get:
Next we fill in the
T-tetromino in the top left, which is easy because it can't go into the 3rd column. For the L above it in the 2nd column, we can just make a start because we don't know whether it's three wide and two high or two wide and three high. (Edit: actually with the updated puzzle we know this for sure, but I'm going to continue with my original solution.)
Turns out that
T-tetrominoes are also easy to solve. We've got part of one in the bottom row, which can only be in the 9th column. That T-tetromino can't go anywhere to the left, so there's only one possible orientation. In the 10th column we've got part of an S below that T, so there's only one possible orientation for this S-tetromino. Then the remainder of the bottom three rows is easy to fill:
Considering again T-tetrominoes:
there must be part of one in the last column, which can only be in the 4th row, and then there's only one possible orientation for that T-tetromino. The next thing to the left of that must be an S-tetromino, which can't now be in the 10th or 11th columns, so we shade a few cells grey. Then a couple of L-tetrominoes are easy to place and we've finished the top right:
S-tetrominoes. There must be part of one in the 9th column. The 8th column must contain two different S-tetrominoes, one of which extends to the right and the other to the left. The 7th column only contains one S-tetromino, so the one that's in the 9th column must stop at the 8th. Then there's only one possible orientation for both S-tetrominoes:
Then there's just two L-tetrominoes left to finish, which gives the two different possibilities shown at the top.