I think this coloring works. Color each row 0 1 2 0 1 2 0 (so the 1st column is all 0, the 2nd is all 1, etc.). This gives you 21 zeros, 14 ones, 14 twos. A straight tile covers three the same (if it's vertical) or one of each (if it's horizontal). If we say it covers $a$ zeros, $b$ ones, and $c$ twos, then in either case $a\equiv b\equiv c$, modulo 3. It follows that the four spaces left after you place the 15 straight tiles are 4 0 0 (that is, 4 zeros, no ones, no twos), or 2 1 1 or 0 2 2. Now the L has to cover some permutation of 2 1 0, and the uncovered square has to be some permutation of 1 0 0. Some of these work, e.g., 2 1 0 plus 0 0 1 gives 2 1 1. But there is no way to add 1 0 0 to any permutation of 2 1 0 to get any of 4 0 0, 2 1 1, or 0 2 2, so the uncovered square can't be a zero, that is, it can't be in the 1st, 4th, or 7th column.
Applying the transpose, it can't be in the 1st, 4th, or 7th row, either. That leaves only 12 places where it can be. By symmetries of the square, these 12 places are of only three types, so if you can find a tiling leaving each of these three types uncovered, you're done. Ross has found a tiling leaving what I'd call $(2,2)$ uncovered; now it remains to do $(2,3)$ and $(3,3)$.
Draw the $2\times n$, for some reasonable number $n$. We will concentrate on the left end of your picture.
The two squares at the end could be filled with two $1\times 1$ tiles. The rest can be done in $t(n-1)$ ways.
There could be one $1\times 1$ tile at the left end, on the top or on the bottom. The remaining square can be filled by an L in $1$ way, and the rest of the $2\times n$ in $t(n-2)$ ways, for a total of $2t(n-2)$.
Or else the there could be an $L$ filling both squares on the left ($2$ ways), with the remaining square filled by a $1\times 1$. That again gives $2t(n-2)$.
Or else we can have an L on the left filling both squares, and an interlocking L. That gives $2t(n-3)$ ways of filling the rest.
The recurrence is therefore
$$t(n)=t(n-1)+4t(n-2)+2t(n-3).$$
For completeness, one should give initial conditions $t(0)$, $t(1)$, $t(2)$.
Best Answer
That is correct so far for the base cases.
Now, suppose for our induction hypothesis that you know how to count all patterns of length strictly less than $n$ for some $n\geq 4$. We want to figure out how many there are for length $n$.
Consider the last column.
Case 1: Both spaces are $1\times 1$ tiles
Case 2: ...
Case 3: ...