If $A$ is a rotation matrix, then $||Ax||=||x||$. Attempt:
Let $$
A =
\begin{bmatrix} 
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}, 
x = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
$$
Then,
$$Ax =  
\begin{bmatrix} 
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix} \cdot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \cos\theta \cdot x_1  -\sin\theta\cdot x_2 \\ \sin\theta \cdot x_1  +\cos\theta\cdot x_2 \end{bmatrix} $$
$$
\begin{align}
||Ax|| &= \sqrt{(\cos\theta \cdot x_1  -\sin\theta\cdot x_2)^2 + (\sin\theta \cdot x_1  +\cos\theta\cdot x_2)^2} \\
&=\sqrt{(\cos^2\theta \cdot x^2_1 -2\cos\theta \cdot x_1\sin\theta\cdot x_2+ \sin^2\theta\cdot x^2_2) + (\sin^2\theta \cdot x^2_1 +2\sin\theta \cdot x_1\cos\theta\cdot x_2+ \cos^2\theta\cdot x^2_2)} \\
&=\sqrt{x^2_1+x^2_2}
\end{align}
$$

However, my text seems to suggest that this is only true for $0\leq\theta\leq\pi$ here:

So where in my attempt did I go wrong?

You didn't go wrong anywhere. You showed what the problem wanted you to show, and then some.
Note that some times the restrictions given in problems and exercises are necessary, and thus not using them means you're wrong somewhere. So you are right to be sceptical. But in this specific case the restriction $0\leq \theta\leq\pi$ is completely superfluous, so there is no issue.
