I would like to contribute here another proof I think interesting.
We use here one simple lemma:
Given two circles (O1), (O2) intersect each other at two points T, U. Any line L passing one intersection point, for example T and intersects (O1), (O2) again at V, W. Suppose Z is midpoint of VW. The result: Z is on circle (O) centered at midpoint O of O1O2 passing T, U.
It is can be proved by draw three perpendiculars from O, O1, O2 to L.
We can apply this lemma to "two equilateral triangles" by drawing two circles (O1), (O2) as circumcircles of two equilateral triangles. After that, all our seven points are midpoints of one line passing one of two intesections of (O1), (O2). Midpoints of XB, YB are two special cases when one end of segment is also intersection point.

Best regards,
Bui Quang Tuan

Keep me updated.

I have send one message before with more two concyclic points but I still don't see it. In any case I resend it here with some more detail information:
If S' = midpoint of XA, T' = midpont of YC
and as your notations in applet:
S = midpoint of AB
T = midpoint of BC
Q = midpoint of AC
Suppose
V = intersection of S'Q and SP
W = intersection of T'Q and TN
then S'Q//XC and SP//AY therefore angle(PVQ) = 60. It means V, B, Q, P are concyclic.
Similarly with W then we have more two concyclic points V and W.
Now we have total ten concyclic points.

We can also use my proof method for two points V, W: they are also two another special cases. In fact:
- VM is tangent line of circumcircle of YBC at M
- WM is tangent line of circumcircle of XAB at M

I hope that we can find one complete nice elemetary proof for generalized fact when XAB, YBC are two similar triangles. In this case we still have ten points on one ellipse. This ellipse is Steiner circumellipse (St) of triangle BKL, here:
K, L is midpoints of AY, CX respectively. Our two circumcircles of XAB, YBC now are two Steiner circumellipses (St1), (St2) of triangles XAB, YBC respectively.
The fact we use for proof is still true:
Any line passing through one intersection of (St1), (St2) will cut three ellipse (St), (St1), (St2) again at three points and point on (St) is midpoint of two other points.

Best regards,
Bui Quang Tuan

By using property: midpoint of crossing segment of two circumcircles (O1) = XAB, (O2) = YBC is on circle (O) we can create a lot of concyclic points. Of course, we only choose points with interesting properties and interesting construction.
Concyclic point may be one alone (for example centroid U or intersection M). Some points are pair (for example N and P)
I suggest one more pair of concyclic points:
X* = intersection of XS with (O1) then X*, B, T' are collinear
Line NO intersects this collinear line X*T' at N' (O is center of our concyclic circle or midpoint of UB).
Easy to show that N' are on circle (O) and N' is reflection of N in center O.
Similarly we can construct P' = reflection of P in center O.
P' and N' also can be accepted as concyclic points.

I think we can find many more concyclic points. Some of them are very easy to prove, other are very hard.
The configuration with one segment, one point on the segment, two equilateral triangles can generate a lot of concyclic points are very similar with Archimedean Arbelos. In the arbelos we can construct a lot of Archimedean circles. In our configuration, we can construct a lot of concyclic points. Of course our configuration can not be interesting as arbelos.
As with arbelos, we can collect all our concyclic points in one catalogue, each point with its proof and its interesting properties. In this case, we must choose proper nomenclature (notations) for concyclic points such as P1, P2,... and may be name for them.
It is can be interesting play for learning how to discover geometry fact and then prove it.

What about generalized fact, I think it is very easy: we can always choose one parallel projection (in 3 dimension space) from one plane to another plane to get equilateral triangles from similar triangles. So what is true for equilateral triangles then it is true for general case and vice versa.

Best regards,
Bui Quang Tuan