Solution: First we prove a lemma:
Lemma 1. Given $\triangle ABC$. Let D, E, F satisfy that $\{AE, AF\}, \{BF, BD\}, \{CD, CE\}$ are respectively the isogonal lines of $\angle BAC, \angle CBA, \angle ACB$. Then AD, BE, CF are concurrent. As shown in the figure:

Proof of Lemma 1. Let AD, BE, CF meet $\odot(BDC), \odot(CEA), \odot(AFB)$ again at X, Y, Z respectively. From
$$\angle BYC = \angle EAC = \angle BAF = \angle BZC$$
we get that B, C, Y, Z lie on a common circle $\Omega_A$. Similarly we obtain that C, A, Z, X lie on a common circle $\Omega_B$, and A, B, X, Y lie on a common circle $\Omega_C$. Hence AD, BE, CF are concurrent at the radical center of $\Omega_A, \Omega_B, \Omega_C$. This completes the proof.

Corollary 1. Given △ABC, let D, E, F be the feet of the perpendiculars from A, B, C to BC, CA, AB respectively. Suppose points X, Y, Z satisfy that \{DY, DZ\}, \{EZ, EX\}, \{FX, FY\} are respectively the isogonal lines of ∠EDF, ∠FED, ∠DFE. Then AX, BY, CZ are concurrent.
Proof of Corollary 1. Let X' satisfy △AEF ∪ X' ~ △ABC ∪ X' and similarly define Y', Z'. Then \{AY', AZ'\}, \{BZ', BX'\}, \{CX', CY'\} are respectively the isogonal lines of ∠BAC, ∠CBA, ∠ACB. Hence by the lemma, AX', BY', CZ' are concurrent at a point T, so AX, BY, CZ are concurrent at the isogonal conjugate of T with respect to △ABC. This completes the proof.

Returning to the original problem, let $P$ be the intersection point of $\odot(ECA)$ and $\odot(FAB)$ other than $A$. Then
$$\begin{align*} \angle BPC &= \angle BPA + \angle APC = \angle BFA + \angle AEC \\ &= \angle BCD + \angle DBC = \angle BDC. \end{align*}$$
Hence $P \in \odot(DBC)$. From
$$\angle APD = \angle APB + \angle BPD = \angle AFB + \angle BCD = 0$$
we get $P \in AD$, and similarly $P$ lies on $BE, CF$.

Let $I_a, I_b, I_c$ be the excenters of △ABC opposite A, B, C respectively, and let Q be the isogonal conjugate of P with respect to △ABC. By the corollary of the lemma (applied to △$II_bI_c$ with $Q, E, F$), $IQ, I_bE, I_cF$ are concurrent. Similarly, $IQ, I_cF, I_aD$ are concurrent, and $IQ, I_aD, I_bE$ are concurrent. Hence $IQ, I_aD, I_bE, I_cF$ are concurrent at a point S.

From
$$\angle ABQ = \angle PBC = \angle ADC$$
we get △ABQ ~ △ADC, so $AI \cdot AI_a = AB \cdot AC = AD \cdot AQ$, and hence
$$\triangle AIQ \sim \triangle ADI_a \Rightarrow \angle AIS = \angle ADS$$
that is $S \in \odot(AID)$. Similarly, $S \in \odot(BIE), \odot(CIF)$. Hence $\odot(AID), \odot(BIE), \odot(CIF)$ are coaxial, completing the proof. See also the figure below: