There is only 1 solution modulo $37$.
The first four equations can be rewritten as $\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, where the entries are taken modulo $37$. The last equation means $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = 1$. Therefore, we need to count the number of $2 \times 2$ matrices $A$ with entries in $\mathbb{F}_{37}$ such that $A^2 = A$ and $\det A = 1$.
Since $\det A = 1 \ne 0$, $A$ is invertible. Thus, we multiply both sides of $A^2 = A$ by $A^{-1}$ to obtain $A = I$. Clearly, the identity matrix $I$ satisfies the conditions. So $A = I$ is the only solution. Equivalently, this means $a \equiv d \equiv 1 \pmod{37}$ and $b \equiv c \equiv 0 \pmod{37}$.