So the basic thing I started with was equating the diagonal and the other other two entries of matrix $A$ respectively which gives, $$a=\pm d, \\b=c$$ But there is this matrix $\begin{bmatrix}0 & 2\\ \frac12 & 0\end{bmatrix}$ who's square also equals $I$.
So, what error did I make? And what will the method to obtain the complete solution set be?
I don't know how you got the condition on the entries of the matrix. There are infinitely many such matrices. To see this note that $A^2=I$ implies that the minimal polynomial of the matrix is $(x-1), (x+1)$ or $(x+1)(x-1)$ since the minimal polynomial divides the annihilating polynomial $x^2-1$. This means that A is diagonalisable and hence has the form $$ A=I,\quad A=-I,\quad A=S\begin{pmatrix}1&0\\0&-1\end{pmatrix}S^{-1} $$ for any invertible matrix $S$. All these choices satisfy $A^2=I$.