This isn't a matter of real versus complex geometry -- the equations you give aren't enough to encode Morley's theorem over the reals. Let $z_1 z_2 z_3$ be our original triangle; I will use indices that are cyclic modulo $3$. Let $\ell^0_i$ be the trisector of the angle at $z_i$ which is closer to $z_{i-1}$ and let $m^0_i$ be the trisector which is closer to $z_i$. Let $\ell^1_i$ and $\ell^2_i$ be the rotations of $\ell^0_i$ by $120^{\circ}$ and $240^{\circ}$, and define $m^1_i$ and $m^2_i$ likewise.
For $r=1$, $2$ and $3$, let $(\delta_r, \epsilon_r)$ equal to one of $(0,0)$, $(1,2)$ or $(2,1)$. Let $c_r$ be the intersection of $m^{\epsilon_r}_r$ and $\ell^{\delta_{r+1}}_{r+1}$. Then $\angle z_r z_{r+1} c_r = \angle c_r z_{r+1} c_{r+1} = \angle c_{r+1} z_{r+1} z_{r+2}$ so your equations hold for $(x_r, y_r) = z_r$ and $(a_r, b_r) = c_r$.
Of the $81$ possible choices for $(\delta_r, \epsilon_r)$, I believe that the triangle $c_1 c_2 c_3$ is equilateral only for $54$ of them. I am basing this on Connes' proof of Morley's theorem, where a key hypothesis is that $$\sum_{r=1}^3 \angle c_{r-1} z_{r} c_{r} = \quad (\ast)$$ is $\pm 60$, not $180$. pm 60^{\circ} \ \mbox{not} \ 180^{\circ}.$$Connes' rotation g_r is a rotation around z_r by angle 2 \angle c_{r-1} z_{r} c_{r}, so what he says is that g_1 g_2 g_3 should be a nontrivial rotation. It's easy enough to encode this condition algebraically. (If you have trouble with it I'll write more, but I bet you won't.) Once you do that, I think it should be easy to convert Connes proof into the sort of proof you are looking for. You'll also need to impose that certain things are nonzero, because Connes divides at times. 1 This isn't a matter of real versus complex geometry -- the equations you give aren't enough to encode Morley's theorem over the reals. Let z_1 z_2 z_3 be our original triangle; I will use indices that are cyclic modulo 3. Let \ell^0_i be the trisector of the angle at z_i which is closer to z_{i-1} and let m^0_i be the trisector which is closer to z_i. Let \ell^1_i and \ell^2_i be the rotations of \ell^0_i by 120^{\circ} and 240^{\circ}, and define m^1_i and m^2_i likewise. For r=1, 2 and 3, let (\delta_r, \epsilon_r) equal to one of (0,0), (1,2) or (2,1). Let c_r be the intersection of m^{\epsilon_r}_r and \ell^{\delta_{r+1}}_{r+1}. Then \angle z_r z_{r+1} c_r = \angle c_r z_{r+1} c_{r+1} = \angle c_{r+1} z_{r+1} z_{r+2} so your equations hold for (x_r, y_r) = z_r and (a_r, b_r) = c_r. Of the 81 possible choices for (\delta_r, \epsilon_r), I believe that the triangle c_1 c_2 c_3 is equilateral only for 54 of them. I am basing this on Connes' proof of Morley's theorem, where a key hypothesis is that$$\sum_{r=1}^3 \angle c_{r-1} z_{r} c_{r} \quad (\ast) is $\pm 60$, not $180$. Connes' rotation $g_r$ is a rotation around $z_r$ by angle $2 \angle c_{r-1} z_{r} c_{r}$, so what he says is that $g_1 g_2 g_3$ should be a nontrivial rotation.