(Elaborating on Did's comment:) You already got that for
$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$
$A^2 = I$ is equivalent to the four equations
$$
\begin{aligned}
a^2+bc &=1 \\
(a+d)b &=0 \\
(a+d)c &= 0 \\
d^2 + bc &= 1
\end{aligned}
$$
Now distinguish two cases:
Case 1: $a+d \ne 0$. Then $b=c=0$ must hold. It follows that $a^2 = d^2 =1$ and therefore $a = d = \pm 1$. The solutions in this case
are
$$
A = I \text{ or } A = -I \, .
$$
Case 2: $a+d = 0$. Then $a^2 + bc=1$ must hold, i.e. the
solutions are
$$
A=\begin{pmatrix} a & b \\ c & -a \end{pmatrix}
\text{ for any $a, b, c$ such that $a^2 + bc=1$} \, .
$$