To follow up on Terry's comment (actually, I did the computation before seeing it, but whatever): in the parabolic case, where the matrix has the form

$$A = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix} \begin{pmatrix} 1 & x \\ 0 & 1\end{pmatrix} \begin{pmatrix} a & b \\ c & d\end{pmatrix},$$
Then 

$$A^n = \begin{pmatrix}1 + c d n x & d^2 n x\\ - c^2 n x & 1 - c d n x\end{pmatrix}.$$
So, if $c \in \mathbb{R},$ then $x < 0$ (from the bottom left), and so $c < 0$ from the bottom right, and we are fine (that is, if $|c d x| > 1,$ then any $a, b$ works). Since $x$ is real, it is clear from the bottom left that $c$ is also, so this finishes the parabolic case.

In the hyperbolic case,
$$A = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix} \begin{pmatrix} x & 0 \\ 0 & 1/x\end{pmatrix} \begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ where $x>1.$

and $$A^n = \begin{pmatrix}=b c x^{-n} + a d x^n & b d x^{-n}(-1 + x^{2 n})\\ a c(x^{-n}+ x^n) & ad x^{-n} - b c x^n\end{pmatrix}.$$ For $n \gg 1,$ we have
$$A^n \sim x^n \begin{pmatrix} ad & b d\\ -ac & -bc\end{pmatrix},$$ which implies that either $a < 0, b>0, c> 0, d<0,$ or $a > 0, b< 0, c<0, d> 0.$ Neither of which is compatible with the base case $n=1$ satisfying your condition.

In the elliptic case, the eigenvalues are either roots of unity, so $A^n = I,$ infinitely often, or not, in which case $A^n$ is very close to $I$ infinitely often, so either way, your condition fails.