The matrix $A$ defines a linear map $\tilde A:\ X\to X$ where $X:={\mathbb R}^2$. To be exact: $A$ is the matrix of $\tilde A$ with respect to the standard basis $\bigl((1,0),(0,1)\bigr)$ of ${\mathbb R}^2$.
(a) Assume $\lambda\in{\mathbb R}$ is a real eigenvalue of $\tilde A$. Then there is a nonzero vector ${\bf e}\in X$ with $A{\bf e}=\lambda{\bf e}$ and therefore $A^2{\bf e}=A\bigl(A{\bf e}\bigr)=\lambda^2{\bf e}$. On the other hand, by assumption on $A$ we know that $A^2{\bf e}=-{\bf e}$, and as ${\bf e}\ne{\bf 0}$ we necessarily would have $\lambda^2=-1$, which is impossible for a real $\lambda$.
(b) Given any ${\bf v}\ne{\bf 0}$ the vector $A{\bf v}$ cannot be a scalar multiple of ${\bf v}$, by (a). It follows that the two vectors ${\bf v}$ and $A{\bf v}$ are linearly independent, and as $X$ is two-dimensional, they form a basis of $X$.
(c) Choose any ${\bf v}\ne{\bf 0}$ and use ${\bf e}_1:={\bf v}$, ${\bf e}_2:=A{\bf v}$ as basis of $X$. Then $A{\bf e}_1={\bf e}_2\ ,\qquad A{\bf e}_2=A^2{\bf e}_1=-{\bf e}_1\ .$ This means that with respect to the basis $({\bf e}_1,{\bf e}_2)$ of $X$ the map $\tilde A$ has matrix $\left[\matrix{0&-1\cr 1&0\cr}\right]=:B\quad.$ By general principles of linear algebra it follows that the matrices $A$ and $B$ are similar.