Let $O$ be the center of $\omega$, and let $M = \overline{PQ} \cap \overline{AB}$ and $N = \overline{PQ} \cap \overline{AC}$ be the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. Refer to the diagram below.

The main idea is to prove two key claims involving $O$, which imply the result:
(i) quadrilaterals $AOE_1F_1$ and $AOE_2F_2$ are cyclic (giving the radical axis is $\overline{AO}$),
(ii) $\triangle OE_1F_1 \cong \triangle OE_2F_2$ (giving the congruence of the circles).

We first note that (i) and (ii) are equivalent. Indeed, because $OP = OQ$, (ii) is equivalent to just the similarity $\triangle OE_1F_1 \sim \triangle OE_2F_2$, and then by the spiral similarity lemma (or even just angle chasing) we have (i) $\iff$ (ii).

Proof of (i) by angle chasing

Note that
$$\angle F_2E_2O = \angle QE_2O = \angle QNO = \angle MNO = \angle MAO = \angle F_2AO$$
and hence $E_2OAF_2$ is cyclic. Similarly, $E_1OAF_1$ is cyclic.

Proof of (i) by Simson lines

Since $P, M, N$ are collinear, we see that $\overline{PMN}$ is the Simson line of $O$ with respect to $\triangle AE_1F_1$.

Proof of (ii) by butterfly theorem

By BUTTERFLY THEOREM on the three chords $\overline{AC}$, $\overline{PQ}$, $\overline{PQ}$, it follows that $E_1N = NE_2$. Thus
$$E_1P = \sqrt{E_1A \cdot E_1C} = \sqrt{E_2A \cdot E_2C} = E_2P.$$
But also $OP = OQ$ and hence $\triangle OPE_1 \cong \triangle OQE_2$. Similarly for the other pair.

Proof of (ii) by projective geometry

Let $T = \overline{PP} \cap \overline{QQ}$. Let $S$ be on $\overline{PQ}$ with $\overline{ST} \parallel \overline{AC}$; then $\overline{TS} \perp \overline{ON}$, and it follows $\overline{ST}$ is the polar of $N$ (it passes through $T$ by La Hire).
Now,
$$-1 = (PQ; NT) \stackrel{T}{\cong} (E_1E_2; N\infty)$$
with $\infty = \overline{AC} \cap \overline{ST}$ the point at infinity. Hence $E_1N = NE_2$ and we can proceed as in the previous solution.

Proof of (ii) by complex numbers

We will give using complex numbers on $\triangle ABC$ a proof that $|E_1P| = |E_2Q|$.
We place $APBCQ$ on the unit circle. Since $\overline{PQ} \parallel \overline{BC}$, we have $pq = bc$. Also, the midpoint of $\overline{AB}$ lies on $\overline{PQ}$, so
$$\begin{aligned} p+q &= \frac{a+b}{2} + \sqrt{\frac{a+b}{2}} \cdot pq \\ &= \frac{a+b}{2} + \frac{a+b}{2ab} \cdot bc \\ &= \frac{a(a+b)}{2a} + \frac{c(a+b)}{2a} \\ &= \frac{(a+b)(a+c)}{2a}. \end{aligned}$$

Now,
$$\begin{aligned} p - e_1 &= p - \frac{pp(a+c) - ac(p+p)}{pp - ac} \\ &= \frac{p(p^2 - p(a+c) + ac)}{pp - ac} = \frac{(p-a)(p-c)}{p^2 - ac} \\ |PE_1|^2 &= (p - e_1) \cdot \overline{p - e_1} = \frac{(p-a)(p-c)}{p^2 - ac} \cdot \frac{\left(\frac{1}{p} - \frac{1}{a}\right)\left(\frac{1}{p} - \frac{1}{c}\right)}{\frac{1}{p^2} - \frac{1}{ac}} \\ &= -\frac{(p-a)^2(p-c)^2}{(p^2 - ac)^2}. \end{aligned}$$
Similarly,
$$|QE_2|^2 = -\frac{(q-a)^2(q-c)^2}{(q^2-ac)^2}.$$
But actually, we claim that
$$\frac{(p-a)(p-c)}{p^2-ac} = \frac{(q-a)(q-c)}{(q^2-ac)^2}.$$
One calculates
$$(p-a)(p-c)(q^2-ac) = p^2q^2 - pq^2a - pq^2c + q^2ac - p^2ac + pa^2c + pac^2 - (ac)^2$$
Thus $(p-a)(p-c)(q^2-ac) - (q-a)(q-c)(p^2-ac)$ is equal to
$$\begin{aligned} & -(a+c)(pq)(q-p) + (q^2-p^2)ac - (p^2-q^2)ac + ac(a+c)(p-q) \\ &= (p-q) \left[ (a+c)pq - 2(p+q)ac + ac(a+c) \right] \\ &= (p-q) \left[ (a+c)bc - 2 \cdot \frac{(a+b)(a+c)}{2a} \cdot ac + ac(a+c) \right] \\ &= (p-q)(a+c) [bc - c(a+b) + ac] = 0. \end{aligned}$$
This proves $|E_1P| = |E_2Q|$. Together with the similar $|F_1P| = |F_2Q|$, we have proved (ii).

Remark. The assumption that $\triangle ABC$ is acute is not necessary; it is only present to ensure that $P$ lies on segment $E_1F_1$ and $Q$ lies on segment $E_2F_2$, which may be helpful for contestants. The argument presented above is valid in all configurations. When one of $\angle B$ and $\angle C$ is a right angle, some of the points $E_1, F_1, E_2, F_2$ lie at infinity; when one of them is obtuse, both $P$ and $Q$ lie outside segments $E_1F_1$ and $E_2F_2$ respectively.