If $A,B$ have eigenvalues $\pm1$, what can I say about the eigenvalues of $AB,BA$?

If you are talking about the relationship between the eigenvalues of $AB$ and $BA$, there is a nice and well-known result, namely, $AB$ and $BA$ have identical spectra.

However, if you are talking about the relationship between the eigenvalues of $A,B$ and $AB$, there is only one thing that one can say:

• the product of the eigenvalues of $AB$ is equal to the product of all eigenvalues in $A$ and $B$; this is because $\det(AB)=\det(A)\det(B)$.

Nothing further can be said without additional information. To illustrate, suppose $n=2$ and the spectra of both $A$ and $B$ are $\{1,-1\}$. The determinantal constraint in the bullet point above dictates that the spectrum of $AB$ must be $\{\lambda,\frac1{\lambda}\}$ for some nonzero $\lambda$. Now, is $\{\lambda,\frac1{\lambda}\}$ really a possible spectrum of $AB$ for every $\lambda\ne0$? The answer is yes. Let $2t=\lambda+\frac1{\lambda}$ and $$ A=\pmatrix{1&0\\ 0&-1},\ B=\pmatrix{t&1\\ 1-t^2&-t}, \ AB=\pmatrix{t&1\\ t^2-1&t}. $$ You may verify that $\operatorname{trace}(B)=0,\,\det(B)=-1,\,\operatorname{trace}(AB)=2t,\,\det(AB)=1$ and hence the characteristic polynomials of $B$ and $AB$ are respectively $x^2-1$ and $x^2-2tx+1=(x-\lambda)(x-\frac1{\lambda})$. Hence the spectrum of $B$ is indeed $\{1,-1\}$ and the spectrum of $AB$ is $\{\lambda,\frac1{\lambda}\}$. In other words, apart from the determinantal constraint mentioned in the bullet point above, the eigenvalues of $AB$ can be pretty much anything.

In general, you can almost never predict the eigenvalues of a product based on the eigenvalues of the matrices you are multiplying together. For example, consider the matrices $$A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \qquad B = \begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix} \qquad C = \begin{bmatrix} -1 & 0 \\ 1 & 1\end{bmatrix}$$ which all have eigenvalues $\pm 1$. Then $$AB = \begin{bmatrix} -1 & 1 \\ 0 & -1 \end{bmatrix}$$ has eigenvalue $-1$, $A^2 = I$ has eigenvalue $1$, and $$BC = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$$ does not have $\pm 1$ as an eigenvalue.

The only glaring exception to this is $0$. If $0$ is an eigenvalue of either $A$ or $B$ then it is also an eigenvalue of $AB$. This is because a matrix is singular if and only if it has $0$ as an eigenvalue.