I have been given a set of $2 \times 2$ matrices, with determinant $1$, Now I have been asked to show that this set forms a group under multiplication (which I have shown) and hence also to find elements of this group with order $2$. This is the part whose solution I want to verify.
My solution:
$$A= \begin{bmatrix}a&b\\c&d\end{bmatrix}$$
Above is a $2 \times 2$ matrix (called $A$) with determinant $1$, i.e. $ad-cb =1$. For any matrix to have order $2$, it must be it's own inverse. The general form of the inverse of this kind of matrices is
$$A^{-1}= \begin{bmatrix}d&-b\\-c&a\end{bmatrix}.$$
This is true since the determinant is $1$. Now for $A$ to be equal to it's own inverse, we must have $a=d$, $c=-c$, $b=-b$. Therefore we must have $c=0$, $b=0$, and $a=d$.
Now the determinant of $A^{-1}$ must also be $1$ (since $\det(A^{-1})=1/\det(A) =1$); hence $ad=1$, since $a=d$. Therefore we must have $a=+1$ or $a=-1$.
Thus the only elements of the group with order $2$ must be $I$, or $-I$, where $I$ is the identity matrix. That's my proposed solution.
Your solution so far mostly makes sense, with one caveat.
What you have shown is that, if $A = A^{-1}$ and a unit determinant is forced, then $A= \pm I$.
Is $I$ or $-I$ an element of order $2$ in this set of matrices? Whether they are still needs to be verified.
Remember that the order of an element $g$ is the minimum $n$ such that $g^n = 1$ (for $1$ the identity of your group $G$).