2×2 matrices without real eigenvalues

Idea. The sign of the lower-left entry $c$ is an invariant, and each sign class is described by explicit inequalities.

Step 1: description. The characteristic polynomial of $A$ is $x^2-(a+d)x+(ad-bc)$, with discriminant
$$\Delta=(a+d)^2-4(ad-bc)=(a-d)^2+4bc.$$
$A\in\mathcal{E}$ iff $\Delta<0$, i.e.
$$(a-d)^2<-4bc.$$

Step 2: sign of $c$. The left side is $\ge0$, so $-4bc>0$, i.e. $bc<0$. In particular $c\ne0$ (and $b\neq0$, with $b$ of sign opposite to $c$). So $\mathcal{E}=\mathcal{E}_+\sqcup\mathcal{E}_-$ with $\mathcal{E}_\pm=\mathcal{E}\cap\{\pm c>0\}$. Both are open (intersections of the open set $\{\Delta<0\}$ with open half-spaces), disjoint and non-empty (they contain $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ and its transpose).

Step 3: $\mathcal{E}_+$ is path connected. $A\in\mathcal{E}_+$ iff $c>0$, $b<0$ and $|a-d|<2\sqrt{-bc}$. Use the coordinates $(b,c,a,\delta)$ with $\delta=a-d$ (a linear change of coordinates, hence a homeomorphism of $\mathbb{R}^4$):
$$\mathcal{E}_+\cong\{(b,c,a,\delta):b<0<c,\ a\in\mathbb{R},\ |\delta|<2\sqrt{-bc}\}.$$
The base $\{(b,c):b<0<c\}$ is convex (a quadrant). Over each point of the base the fibre $\{(a,\delta)\}$ is $\mathbb{R}\times(-2\sqrt{-bc},2\sqrt{-bc})$, convex, and the bounds depend continuously on $(b,c)$. To join two points $(b_0,c_0,a_0,\delta_0)$ and $(b_1,c_1,a_1,\delta_1)$, first shrink $\delta$ linearly to $0$ (the inequality $|\delta|<2\sqrt{-bc}$ remains true), then move $(b,c)$ along the segment while keeping $\delta=0$ and moving $a$ linearly, then expand $\delta$ linearly to $\delta_1$ (allowed because the constraint at the new $(b_1,c_1)$ is $|\delta_1|<2\sqrt{-b_1c_1}$).

Step 4: same for $\mathcal{E}_-$, with $c<0<b$. It is homeomorphic to $\mathcal{E}_+$ via $A\mapsto A^T$ (which swaps $b$ and $c$).

Conclusion. $\mathcal{E}$ has exactly two path components, $\mathcal{E}_+$ and $\mathcal{E}_-$.

Remark. Geometrically, $A\in\mathcal{E}_\pm$ determines an *orientation*: $A$ acts like a rotation in the plane, counterclockwise for $c>0$ and clockwise for $c<0$ (the sign of $c$ is the sign of $\det\big(v,Av\big)$ for any vector $v\ne0$). This invariant has no analogue for matrices with real eigenvalues.