My first post was close so i tried it again:
This is a question from a university test in wich i have half of the score, so i post here the development i made for you to comment possible errors:
Be $M \in M_{2}[\mathbb{R}]$,$M=\left(\begin{matrix} a & b \\ c & d \end{matrix}\right)$ such that $Det(M)=1$ stablish conditions over the values of $a,b,c,d \in \mathbb{R} $ such that $M$ is diagonalizable.
what i do is to take the matrix $ \left(\begin{matrix} a-\lambda & b \\ c & d-\lambda \end{matrix}\right)$ and evaluating the characteristic polynomial $(a-\lambda)(d-\lambda)-bc=\lambda^2 - \lambda(a+d) + 1$ we have two distinc roots if and only if the discriminant $(a+d)^2-4>0$ so if $a+d<-2$ or $a+d >2$ there exist two real distincts eigenvalues and $M$ is diagonalizable.
After that we analyse the case $a+d=2$ or $a+d=-2$ in the first case we have for the characteristic polynomial $\lambda^2-2\lambda+1=(\lambda+1)^2=0$ so an eigenvalue $\lambda=-1$ of multiplicity 2 , for the second case the characteristic polynomial $(\lambda-1)^2=0$ give us a second eigenvalue $\lambda=1$ again with miltiplicity 2.
reemplazing in the orginal matrix we have, for the value $\lambda=1$ $$ \left(\begin{matrix} a-1 & b \\ c & d-1 \end{matrix}\right) \left(\begin{matrix} x_{1} \\ x_{2} \\ \end{matrix}\right) = \left(\begin{matrix} 0 \\ 0 \end{matrix}\right) $$
we have that the geometric multiplicity of the solution is two and so $M$ is diagonalizable only if $0=a-1=b=c=d-1$ and so $M=I$ in an analogous form we find that $a+1=d+1=b=c=0$ permits the diagonalization of $M$ that is $M=-I$. In the final case that $-2<a+d<2$ we have no real eigenvalues. I have tried to redact my answer in the form of my test so i thank your answers.
The analysis is correct. On the other hand, it's perhaps simpler to prove that a $2\times2$ real matrix over the reals is diagonalizable if and only if it has two distinct real eigenvalues or is diagonal.
Indeed, the characteristic polynomial can be written in the form $$ \lambda^2-(a+d)\lambda+(ad-bc) $$ If the polynomial has distinct real roots, then the matrix is diagonalizable. Such condition is $$ (a+d)^2-4(ad-bc)>0 $$ that is $$ (a-d)^2+4bc>0 $$ (in your case $ad-bc=1$). In the case when the discriminant is $0$, the matrix has a single real eigenvalue of algebraic multiplicity $2$ and it is $\lambda=(a+d)/2$ (no need to use the quadratic formula). In order to find the geometric multiplicity, we have to consider $$ \begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix}= \begin{pmatrix} (a-d)/2 & b \\ c & (d-a)/2 \end{pmatrix} $$ and the geometric multiplicity is $2$ if and only if this is the zero matrix, that is, $a=d$ and $b=c=0$. Thus the matrix must be diagonal to begin with. In the case of determinant $1$, the condition is that $a=d=1$ or $a=d=-1$, $b=c=0$.