This is *not* a solution, rather it is a conjectured generalization of the claimed identity which may help in proving it:

The idea is to observe that $x^2$ is a primitive $n$-th root of unity, and to replace the factor $-x$ in $-x^{2j+1}=(-x)(x^2)^j$ with a new variable $z$.

Thus let $\zeta$ be a primitive $n$-th root of unity and $z$ be a variable. Then it seems to be the case that
\begin{equation}
\mathop{\mathrm{Tr}}\left(\prod_{j=0}^{n-1}\begin{pmatrix}  
\zeta^{-j} & \zeta^jz \\
1 & 0
\end{pmatrix}\right)=
\begin{cases}
3(1-\zeta^{n/6})z^{n/3}-1 & \text{if } n\equiv0\pmod{6}\\
3(\zeta z)^{n/3} & \text{if } n\equiv3\pmod{6}\\
-(-1)^n & \text{if } n\not\equiv0,3\pmod{6}
\end{cases}.
\end{equation}
Let $P(z)$ the term on the left hand side. It is clear that $P(z)$ is a polynomial in $z$ of degree at most $n-1$. Cyclically permuting the factors on the left hand side does not change the trace, and neither does conjugating each factor with the same matrix. Set $\delta=\zeta^k$ for some $0\le k\le n-1$. Then
\begin{equation}
\begin{pmatrix}1&0\\0&\delta\end{pmatrix}
\begin{pmatrix}\zeta^{-j} & \zeta^jz \\1 & 0\end{pmatrix}
\begin{pmatrix}1&0\\0&1/\delta\end{pmatrix}
=
\begin{pmatrix}\zeta^{-j} & \zeta^jz/\delta \\\delta & 0\end{pmatrix}=
\delta\begin{pmatrix}\zeta^{-j-k} & \zeta^{j+k}(z\zeta^{-3k}) \\1 & 0\end{pmatrix}.
\end{equation}
This shows $P(z)=P(z\zeta^{3k})$ for all $0\le k\le n-1$.

So if $n\not\equiv0\pmod{3}$, then $P(z)$ must be a constant. One easily computes $P(0)$.

In the other cases, we only know that $P(z)$ is a polynomial in $z^{n/3}$ of degree less than $n$. I believe that a refinement of the argument should settle these cases too.