It is clear that $a_n = 0$ unless $n$ is divisible by 3, so I let
$b_n = a_{3n}$.
According to the OEIS link provided
by Douglas Zare in his comment above, we have (with a short calculation
to obtain the numerator)
$$ \sum_{n=0}^\infty c_n t^n = F(t) := \frac{1 - 2t + t^2}{1 - 5t + 3t^2 - t^3}, $$
where $c_n$ denotes the number of tesselations of a $4 \times 3n$-rectangle
without identifying tesselations in the same orbit of the symmetry group.

Now we have to consider the possible symmetries. It is easy to see that
the number of tesselations that are invariant under the rotation
by 180 degrees is $c_{\lfloor n/2 \rfloor}$ (if $n$ is even, there
can be no tile extending across the axis, so the tesselation is given
by its left half; if $n$ is odd, the middle part of the tesselation has
to consist of four horizontal tiles, and a similar argument applies).

The same is true for tesselations invariant under the reflection
about the vertical axis when $n$ is even. When $n$ is odd, the
middle part can consist of four horizontal tiles or of one
vertical and one horizontal tile (at top or bottom).
Denoting the number of tesselations of a $4 \times 3n$-rectangle
with a vertical $1 \times 3$ rectangle removed at one end by $d_n$,
we obtain $c_{(n-1)/2} + 2 d_{(n+1)/2}$ possibilities.
The generating function of the $d_n$ can be fairly easily obtained
from that for the $c_n$; it is
$$ \sum_{n=0}^\infty d_n t^n = G(t) := \frac{t - t^2}{1 - 5t + 3t^2 - t^3}. $$

There is exactly
one tesselation that has both (and therefore all) symmetries (the one
with all tiles horizontal); it is also the only one that is symmetric
with respect to the reflection about the horizontal axis.

So there is one orbit of size 1
and there are $2(c_{n/2} - 1)$ orbits of size 2 when $n$ is even
and $2(c_{(n-1)/2} - 1 + d_{(n+1)/2})$ when $n$ is odd,
so there must be
$$ b_n = \frac{1}{4}(c_n + \#\text{(orbits of size 2)} + 3)
= \frac{1}{4}(c_n + 1)
+ \frac{1}{2} c_{\lfloor n/2 \rfloor}
\Bigl[+ \frac{1}{2} d_{(n+1)/2} \text{ if $n$ is odd}\Bigr]$$
orbits in total.
Therefore
$$ \sum_{n=0}^\infty b_n t^n
= \frac{1}{4} \Bigl(F(t)+\frac{1}{1-t}\Bigr)
+ \frac{1}{2}\bigl((1 + t) F(t^2) + t^{-1} G(t^2)\bigr), $$
which is
$$ \frac{1-4t-4t^2+20t^3-8t^4-13t^5+10t^6+t^7-3t^8+t^9}{(1-t)(1-5t+3t^2-t^3)(1-5t^2+3t^4-t^6)}. $$