It is given that $ A = \left[ \begin{array}{rrr} 1 & -2 & 0 \\ 1 & 1 & 1 \\ -1 & 3 & -2 \\ \end{array} \right] $ and $ \vec{x} = \left[ \begin{array}{r} x_1 \\ x_2 \\ x_3 \\ \end{array} \right] $.
It is then asked that the homogeneous system $(A-I)\vec{x}=\vec{0}$ be solved and that this result be used to solve $A\vec{x} = \vec{x}$ for $ \vec{x} = \left[ \begin{array}{r} x_1 \\ x_2 \\ x_3 \\ \end{array} \right] $ and $A\vec{x}=-2\vec{x}$.
I'm not really certain what exactly is going on here, but:
I begin $(A-I)\vec{x}=\vec{0} \implies A\vec{x}-I\vec{x}=\vec{0} \implies A\vec{x} = I\vec{x} \implies A\vec{x}=\vec{x}$ which can only be true if $\vec{x}=\vec{0}$ or $A=I$, but indeed we are given $A \neq I \implies \vec{x}=\vec{0}$.
I don't know how terribly helpful this is, or, at least it seems so trivial that it must be wrong.
You are right that your conclusion is flawed.
From $A\mathbf{x}=\mathbf{x}$, it is not possible to conclude that either $A=I$ or $\mathbf{x}=\mathbf{0}$. In fact, the vectors satisfying this property, or more generally $A\mathbf{x}=\lambda\mathbf{x}$, are called eigenvectors with eigenvalue $\lambda$. As an example, you may check that $\mathbf{v}=[1,2]^T$ is an eigenvector of \begin{align*} \left[\begin{array}{cc} 6&-2\\ 0&2 \end{array}\right]. \end{align*} In the given exercise you are probably supposed to realize that solving $A\mathbf{x}=\mathbf{x}$ is equivalent to solving $(A-I)\mathbf{x}=\mathbf{0}$. The solutions are the eigenvectors with eigenvalue $1$.
How may we use this idea to solve $A\mathbf{x}=-2\mathbf{x}$? Are the solutions eigenvectors? If yes, what is the eigenvalue?