Question: I have a hard time visualizing what the actual elements of this ring are. Because it is mod the elements generated by $x^2 - 1$ are polynomials with non-even powers in the ring or not?

Answer: 
The elements of the ring can be thought of as the polynomials $a+bx$ of degree $\le 1$, with the "normal" addition, and multiplication almost normal, except that $x^2$ is replaced by $1$.
Notice that $x^2-1 = (x-1)(x+1)$, and $x+1$ and $x-1$ are both irreducible polynomials in $\mathbb{Z}[x]$. Thus by the Chinese Remainder Theorem we have
\begin{equation}
\mathbb{Z}[x]/\langle x^2 - 1 \rangle \cong \mathbb{Z}[x]/\langle x+1 \rangle \times\mathbb{Z}[x]/\langle x-1 \rangle \cong \mathbb{Z} \times \mathbb{Z}
\end{equation}
So this ring is just integer pairs. Does that answer your question?