How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$?

Find the three cube roots of $1$. Let $a+bi$ be one of those. They are solutions of $x^3=1$. Hence $x^3-1=0$.

Since $1$ is one of the solutions, $x-1$ must be one of the factors, thus: $$ x^3-1 = (x-1)(\cdots\cdots\cdots). $$ Fill in the blanks by doing long division. You should get $$ (x-1)(x^2+x+1). $$ So the equation is $$ (x-1)(x^2+x+1) = 0. $$ That implies $$ x-1=0\quad\text{or}\quad x^2+x+1 = 0. $$ Solve the quadratic equation.

Note that there are also $2\times2$ matrices which are not of the form indicated in the question which solve the problem that $A^3=I$. The general solution to your problem is ($\alpha,\beta \in \mathbb{R}, \beta\neq 0$) $$A =\begin{pmatrix} \alpha & \beta\\ -\beta^{-1}(1+\alpha+\alpha^2) & -(1+\alpha)\end{pmatrix} .$$

The specific examples which correspond to complex numbers $a+ib$ need to fulfill $$\alpha = -(1+\alpha) \quad \text{and} \quad\beta = \beta^{-1}(1+\alpha+\alpha^2)$$ with the solutions $\alpha=-1/2$, $\beta=\pm\sqrt{3}/2$ (corresponding to rotation by $\pm 2\pi/3$).

The matrix $A=\begin{pmatrix} 0&1\\-c&-b\end{pmatrix}$ satisfies $A^2+bA+cI=0$.

So the matrix $\begin{pmatrix} 0&1\\-1&-1\end{pmatrix}$ satisfies $A^3-I=(A-I)(A^2+A+I)=0$