Ex-Skews Me Sir

Color the area as follows with $28$ gray squares and $32$ white squares:

Then there are two types of skew tetronimoes that can be placed on the board - one that covers $1$ white square and $3$ gray squares, and one that covers $3$ white squares and $1$ gray square:

Let $x$ be the number of skew tetronimoes that cover $1$ white square and $3$ gray squares, and let $y$ be the number of skew tetronimoes that cover $3$ white squares and $1$ gray square.

Since each skew tetronimo covers $4$ squares, and since there are $8 \cdot 8 - 4 = 60$ squares, there are $\frac{60}{4} = 15$ total skew tetronimoes, so $x + y = 15$ .

Also, since $28$ gray squares need to be covered, $3x + y = 28$ .

However, $x + y = 15$ and $3x + y = 28$ leads to $(x, y) = (\frac{13}{2}, \frac{17}{2})$ , a non-integer solution, which is not possible.

Therefore, the tiling is not possible .

Label the squares from $1$ to $4$ as follows:

Any rotation of a "Z" skew tetromino (like the green or purple ones in the original diagram) always covers $1,2,3,4$ in some order.

Any rotation of an "S" skew tetromino (like the blue one in the original diagram) always covers two pairs of numbers. Count each type of tile by the set of numbers it covers:

There are $15$ ones and $16$ twos in the grid. This means that $A+2B+2E=15$ and $A+2B+2C=16$

But the first of these implies $A$ is odd, and the second that $A$ is even; contradiction. So no such tiling is possible .