What is the number of distinct elements in $S$?

Allow for these values:

$$A = \begin{pmatrix} \cos \frac{2 \pi}{5} & -\sin \frac{2 \pi}{5} \\ \sin \frac{2 \pi}{5} & \cos \frac{2 \pi}{5} \end{pmatrix} \text{ and } B = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$

Set $S$ as the set of all matrices which can be generated from taking the products of matrices $A$ and $B$--this will work in any order. As an example, the matrix $A^2BA^4B$ is in set $S$. What is the number of distinct elements in $S?$

I realize this question has been asked a couple of times before. However, in here: Number of Distinct Elements in Set of Products of 2 Matrices, the answer I saw was confusing to me as I have not learned those terms yet. Additionally, in here: Geometry in Vectors, geometry was used, but I am looking to utilize an approach that involves vectors/matrices, not through geometrical shapes.

I'm not sure how to solve it that way though, and I could really use some hints. I'm very confused and am not sure how to start.

The matrix $A$ rotates the plane counterclockwise through one-fifth of a circle. Doing that five times brings you back to your starting point. Thus $A^5 = A^0 = I$. The group generated by $A$ alone thus has five members: $A^0=I, A^1, A^2, A^3, A^4$.
The matrix $B$ reflects about the $x$-axis. The $x$-axis is the mirror, so that for example, the two points $\begin{bmatrix} 5 \\ \pm 3 \end{bmatrix}$ are each other's mirror images: $$B \begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} 5 \\ -3 \end{bmatrix} \text{ and } B \begin{bmatrix} 5 \\ -3 \end{bmatrix} = \begin{bmatrix} 5 \\ 3 \end{bmatrix}.$$
Try to show that if you rotate a certain distance counterclockwise and then reflect about the $x$-axis, you get the same result as if you had first reflected about the $x$-axis and that rotated clockwise. From that you can conclude: \begin{align} BA = A^4B & \quad \text{ or } \quad BA^{-4} = A^{-1} B \\ BA^2 = A^3B & \quad \text{ or } \quad BA^{-3} = A^{-2} B \\ BA^3 = A^2B & \quad \text{ or } \quad BA^{-2} = A^{-3} B \\ BA^4 = A^{-1}B & \quad\text{ or } \quad BA^{-1} = A^{-4} B \end{align}
So you have $$\{I,A,A^2, A^3, A^4, B, BA, BA^2, BA^3, BA^4\} = \{I,A,A^2, A^3, A^4, B, AB, A^2B, A^3B, A^4B\}\}.$$