WLOG, say the center of the circle ($O$) is at the origin. Vertices of the pentagon $ABCDE$ are represented by position vectors $\overline{a}, \overline{b}, \overline{c}, \overline{d}$ and $\overline{e}$.
Centroid of $\triangle ABC, \, \overline {g} = \frac{\overline{a} + \overline{b} + \overline{c}}{3}$
Line $DE = \overline{d} - \overline{e}$
As points $A, B, C, D, E$ are concyclic with center at $O$
$|\overline{a}|^2 = |\overline{b}|^2 = |\overline{c}|^2 = |\overline{d}|^2 = |\overline{e}|^2$ ...(i)
If a point $P$ with position vector $\overline{p} \,$ is on the perpendicular line from the centroid of $\triangle ABC$ to the line $DE$,
$(\overline{p}-\overline{g}) \cdot (\overline{d} - \overline{e}) = 0$
Based on (i) one of the ways for the dot product to be zero is
$(\overline{p}-\overline{g}) = n_1 (\overline{d}+\overline{e}) \,$ (you can easily show why $\overline{p} = \overline{g}$ will not give you the concurrent point by symmetry)
$\overline{p}-\overline{g} = \overline{p}-\frac{\overline{a} + \overline{b} + \overline{c}}{3} = n_1 (\overline{d}+\overline{e})$ ...(ii)
Similarly,
$\overline{p}-\frac{\overline{b} + \overline{c} + \overline{d}}{3} = n_2 (\overline{e}+\overline{a})$ ...(iii)
From (ii)-(iii), you get one solution when $n_1 = n_2 = \frac{1}{3}$ and
$\overline {p} = \frac{\overline{a} + \overline{b} + \overline{c} + \overline{d} + \overline{e}}{3}$
Now we need to prove this point is the point of concurrency for other $3$ lines too. So we take the lines from centroids of $\triangle CDE, \triangle DEA, \triangle EAB$ through point $\overline {p}$ and show each of them is perpendicular to the line segment made by other two vertices.
$(\overline{p}- \frac{\overline{c} + \overline{d} + \overline{e}}{3}) \cdot (\overline{a} - \overline{b}) = 0$
$(\overline{p}- \frac{\overline{d} + \overline{e} + \overline{a}}{3}) \cdot (\overline{b} - \overline{c}) = 0$
$(\overline{p}- \frac{\overline{e} + \overline{a} + \overline{b}}{3}) \cdot (\overline{c} - \overline{d}) = 0$
which is easy to show given (i).
After some symbol-bashing in Mathematica, here are barycentric coordinates of key points:
$$\begin{align}
A' &= (0 : a + b - c : a - b + c) \\[0.75em]
A'' := \tfrac12(B'+C') &=
\left(a +\frac{(a-b+c)(a+b-c)}{-a+b+c} : b : c\right) \\[0.75em]
A''' &= (2 a + b + c : b : c ) \\[0.75em]
O_A &= (2 a^3 : a^3 - b^3 + c^3 + a^2 b - a^2 c - 3 a b^2 - a c^2 - b^2 c + b c^2 - 4 a b c \\
&\phantom{(2 a^3\;}\quad : a^3 + b^3 - c^3 - a^2 b + a^2 c - a b^2 - 3 a c^2 + b^2 c -
b c^2 - 4 a b c)
\end{align}$$
The coordinates of the points where, say, $\bigcirc O_A$ (the one passing through incenter $X_1$) meets the side-lines of the circle are a little messy, so I'll omit them. Nevertheless, the six prescribed points are indeed cyclic, and the equation of their common circle has this barycentric form (in $u:v:w$ coordinates):
$$\begin{align}
0 &= u^2 b c (-a + b + c) \cos A + v^2 c a (a - b + c) \cos B + w^2 a b (a + b - c) \cos C \\[0.75em]
&\quad+ v w (-a^3 + b^3 + c^3 - 2 a^2 b - 2 a^2 c - b^2 c - b c^2) \\[0.75em]
&\quad+ w u (-b^3 + c^3 + a^3 - 2 b^2 c - 2 b^2 a - c^2 a - c a^2) \\[0.75em]
&\quad+ u v (-c^3 + a^3 + b^3 - 2 c^2 a - 2 c^2 b - a^2 b - a b^2)
\end{align}$$
Converting to trilinear form (in $\alpha:\beta:\gamma$ coordinates), the equation can be manipulated into this:
$$(\lambda \alpha + \mu\beta+\nu\gamma)(a\alpha+b\beta+c\gamma)+\kappa(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$
where
$$\begin{align}
\lambda &:= (-a + b + c)\cos A \\
\mu &:= (\phantom{-}a-b+c)\cos B\\
\nu &:= (\phantom{-}a+b-c)\cos C\\
\kappa &:= -2(a+b+c)
\end{align}$$
As $\lambda:\mu:\nu$ are the trilinear coordinates for Kimberling center $X(219)$ ("$X(8)$-Ceva Conjugate of $X(55)$"), the circle is the Central Circle of that point. This answers the "alternatively" aspect of OP's question.
I don't know what Kimberling centers may lie on the circle. I don't even know how one would go about searching for them. Someone with more fluency in triangle-center lore may have some insights.
Incidentally, for the $\bigcirc O_A$ passing through $B''$, $B'''$, $C''$, $C'''$, etc, the corresponding six-point circle has trilinear form
$$(\lambda'\alpha+\mu'\beta+\nu'\gamma)(a\alpha+b\beta+c\gamma)+\kappa'(a\beta\gamma+b\gamma\alpha+c\alpha\beta)=0$$
where
$$\begin{align}
\lambda' &:= a (-a + b + c)^2 \\
\mu' &:= b (\phantom{-}a - b + c)^2 \\
\nu' &:= c (\phantom{-}a + b - c)^2 \\
\kappa' &:= -8 a b c
\end{align}$$
This circle is therefore the Central Circle of Kimberling's $X(220)$ ("$X(9)$-Ceva conjugate of $X(55)$").
I have not investigated whether there is a general connection between OP's construction, Ceva-conjugates, and/or $X(55)$.
Best Answer
Problems like this make me want to look for general principles instead of bogging-down in messy specifics. In this case, we have a triangle $\triangle ABC$, say, with circumcenter $O$, and circular-segment-centroids $A'$, $B'$, $C'$ (OP's $A_1$, $B_1$, $C_1$), with $\overline{OA'}$, $\overline{OB'}$, $\overline{OC'}$ bisecting $\angle BOC$, $\angle COA$, $\angle AOB$, respectively; a little calculus tells us the centroids' distances from the circumcenter. (We'll get to that later.)
Stepping back, we see that we have six points $A$, $B$, $C$, $A'$, $B'$, $C'$ arranged about a common center, $O$, and that we know the distances of these point from $O$ and angles determined by these points and $O$. We can establish a condition on those lengths and angles that guarantees the concurrence of $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$. So let's do that.
Define $$ a := |OA| \quad b := |OB| \quad c := |OC| \quad a' := |OA'| \quad b' := |OB'| \quad c' := |OC'| $$ (Note that we're generalizing beyond $O$ being the circumcenter, which would required $a=b=c$. We're also generalizing beyond, say, $\overline{OA'}$ bisecting $\angle BOC$, etc; the general rule turns out to be pretty nice without these assumptions.) We'll also use various angles, $\angle XOY$, taken to be oriented "from" $X$ "to" $Y$; this allows us to write $\angle XOY+\angle YOZ=\angle XOZ$ and $\angle XOY=-\angle YOX$.
Now, let's coordinatize. Abusing notation to define $\operatorname{cis}\theta := (\cos\theta, \sin\theta)$ we can take $$\begin{align} A &:= a \operatorname{cis}0 && A' := a'\operatorname{cis}\angle AOA'\\ B &:= b \operatorname{cis}\angle AOB && B' := b'\operatorname{cis}\angle AOB' \\ C &:= c\operatorname{cis}\angle AOC && C' := c'\operatorname{cis}\angle AOC' \end{align}$$
From here, the process is straightforward, if tedious. (It helps to have a computer algebra system to crunch the symbols.) We determine the equations of the lines $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$, find the intersection of any two, and substitute the intersection into the third. When the dust settles (and barring degeneracies), we get a relation that we can express thusly:
$$\begin{align} 0 &= \phantom{+}a a' \sin\angle AOA'\; \left( b c \sin\angle BOC +c b' \sin\angle COB' +b' c' \sin\angle B'OC' +c' b \sin\angle C'OB \right) \\[4pt] &\phantom{=} +b b'\sin\angle BOB'\; \left( c a \sin\angle COA +a c' \sin\angle AOC' +c' a' \sin\angle C'OA' +a' c \sin\angle A'OC \right) \\[4pt] &\phantom{=}+ c c' \sin\angle COC'\; \left( a b \sin\angle AOB +b a' \sin\angle BOA' +a' b' \sin\angle A'OB' +b' a \sin\angle B'OA \right) \end{align} \tag{$\star$}$$
This may seem a little daunting at first glance, but, glancing again, we notice that every "$\sin\angle XOY$" is multiplied by the corresponding lengths "$x$" and "$y$"; conveniently, each such product is thus twice the (signed) area $|\triangle XOY|$, so that we can write
$$\begin{align} 0 &= \phantom{+}|\triangle AOA'|\; \left( |\triangle BOC|+|\triangle COB'|+|\triangle B'OC'|+|\triangle C'OB| \right) \\[4pt] &\phantom{=} +|\triangle BOB'|\; \left( |\triangle COA|+|\triangle AOC'|+|\triangle C'OA'|+|\triangle A'OC| \right) \\[4pt] &\phantom{=}+ |\triangle COC'|\; \left( |\triangle AOB|+|\triangle BOA'|+|\triangle A'OB'|+|\triangle B'OA| \right) \end{align} \tag{$\star\star$}$$
Even better, each long factor is the sum of the (signed) areas of adjacent triangles that form a quadrilateral; so each factor gives the (signed) area of that quadrilateral. (This interpretation is a bit nuanced in cases where such a quadrilateral is self-intersecting. Be that as it may ...) This gives us this streamlined expression:
$$ |\triangle AOA'|\;|\square BCB'C'| +|\triangle BOB'|\;|\square CAC'A'| +|\triangle COC'|\;|\square ABA'B'| =0 \tag{$\star\star\star$}$$
Pretty nifty! $\square$
Now that we have generalized the problem, let's work our way towards the specifics of OP's ostensible concurrence.
We consider $\triangle ABC$ with interior angles $\alpha := \angle A$, $\beta := \angle B$, $\gamma := \angle C$. Taking $O$ to be the circumcenter and $r$ the circumradius, we have $$a=b=c=r \qquad \angle BOC = 2\alpha \quad \angle COA = 2\beta \quad \angle AOB = 2\gamma $$ With $A'$, $B'$, $C'$ along the bisectors of $\angle BOC$, $\angle COA$, $\angle AOB$, we have $$\angle BOA' = \angle A'OC=\alpha \qquad \angle COB'=\angle B'OA=\beta \qquad \angle AOC'=\angle C'OB=\gamma$$ $$\angle AOA' = 2\gamma+\alpha=\pi-(\beta-\gamma) \qquad \angle BOB' = \pi-(\gamma-\alpha) \qquad \angle COC' = \pi-(\alpha-\beta)$$
Falling back to version $(\star)$ of our concurrence condition, we have $$\begin{align} 0 &= \phantom{+}r a' \sin(\beta-\gamma)\; \left( r^2 \sin2\alpha +r b' \sin\beta +b' c' \sin(\beta+\gamma) +c' r \sin\gamma \right) \\[4pt] &\phantom{=} +r b'\sin(\gamma-\alpha)\; \left( r^2 \sin2\beta +r c' \sin\gamma +c' a' \sin(\gamma+\alpha) +a' r \sin\alpha \right) \\[4pt] &\phantom{=}+ r c' \sin(\alpha-\beta)\; \left( r^2 \sin2\gamma +r a' \sin\alpha +a' b' \sin(\alpha+\beta) +b' r \sin\beta \right) \end{align} \tag{1}$$ Since $\alpha+\beta+\gamma=\pi$ and $r\neq 0$, this simplifies to
$$\begin{align} 0 &= \phantom{+}(r a' - b' c') \sin2\alpha \sin(\beta - \gamma) \\ &\phantom{=}+(r b' - c' a') \sin2\beta \sin(\gamma - \alpha) \\ &\phantom{=}+(r c' - a' b') \sin2\gamma \sin(\alpha - \beta) \end{align} \tag2$$
Note that $(2)$ holds for $A'$, $B'$, $C'$ anywhere along the perpendicular bisectors, so it's still a bit of a generalized result. For OP's circular-segment-centroids, we consult Wikipedia's "List of Centroids" to remind ourselves that $$a' = \frac{4r\sin^3\alpha}{3(2\alpha-\sin2\alpha)} \qquad b' = \frac{4r\sin^3\beta}{3(2\beta-\sin2\beta)} \qquad c' = \frac{4r\sin^3\gamma}{3(2\gamma-\sin2\gamma)} \tag{3}$$
Perhaps-unsurprisingly, upon substituting the values from $(3)$ into $(2)$, the mix of "raw and trigged" angles doesn't simply vanish. For the sake of completeness, here's a version of the resulting concurrence condition
$$\begin{align} &\phantom{=+\,} 3 \sin^3\alpha \sin(\beta-\gamma) (\alpha \sin2\beta\sin2\gamma + 2\beta\gamma \sin2\alpha) \\ &\phantom{=} +3 \sin^3\beta \sin(\gamma-\alpha) (\beta \sin2\gamma\sin2\alpha + 2\gamma\alpha \sin2\beta) \\ &\phantom{=} +3 \sin^3\gamma \sin(\alpha-\beta) (\gamma \sin2\alpha\sin2\beta + 2\alpha\beta \sin2\gamma) \\[6pt] &= 8 \sin\alpha \sin\beta \sin\gamma \left(\begin{array}{l} \phantom{+} \alpha \cos\alpha \sin^2\beta \sin^2\gamma \sin(\beta-\gamma) \\ + \beta \cos\beta \sin^2\gamma \sin^2\alpha \sin(\gamma-\alpha) \\ + \gamma \cos\gamma \sin^2\alpha \sin^2\beta \sin(\alpha-\beta) \\ + \sin\alpha \sin\beta \sin\gamma \sin(\beta-\gamma) \sin(\gamma-\alpha) \sin(\alpha-\beta)\end{array}\right) \end{align}\tag{4}$$
OP's alternative construction, taking $A'$, $B'$, $C'$ to be the centroids of the "other" circular segments, requires the substitutions $\alpha\to\pi-\alpha=\beta+\gamma$, $\beta\to\gamma+\alpha$, $\gamma\to\alpha+\beta$ in $(3)$, along with changing the sign of each of $a'$, $b'$, $c'$ because each centroid lies on the "other side" of $O$. These adjustments cause some minor sign changes in $(4)$, but also introduce more-complicated "raw" angle expressions. The result doesn't appreciably simplify, so I won't bother TeX-ing it up.