Here is a hugeee overkill solution that probably works? We use Forgotten Coaxiality Lemma for this. I'm quite sure there exists a wayyyy easier solution for this ? We keep the point $X$ free on $AC$ and define $Y=BC\cap \odot(CDX)$. Now let $P=AB\cap \odot(CBX),Q=AD\cap \odot(CDX)$. Now $M,N$ be the midpoints of $AB,AQ$ respectively. We redefine $E=\odot(CBX) \cap \odot(CDN)$. Now $R=BC\cap \odot(CDN)$. Finally, $Y'$ be the reflection of $B$ over $R$. $\textbf{CLAIM: }MPND$ is cyclic. $\textbf{PROOF: }$ $$AM\cdot AP=\dfrac{AB \cdot AP}{2}=\dfrac{\operatorname{Pow}_{\odot CPBX}(A)}{2}=\dfrac{AX\cdot AC}{2}=\dfrac{\operatorname{Pow}_{\odot CXDQ}(A)}{2}=\dfrac{AD\cdot AQ}{2}=AD\cdot AN.$$ $\textbf{CLAIM: }\overline{P-E-N}$ are collinear. $\textbf{PROOF: }$ $$\measuredangle CEP=\measuredangle CBP=\measuredangle CBA=\measuredangle CDA=\measuredangle CDN=\measuredangle CEN.$$ $\textbf{CLAIM: }AB\parallel NR\parallel QY'$. $\textbf{PROOF: }$ $$\measuredangle CRN=\measuredangle CDN=\measuredangle CDA=\measuredangle CBA \implies AB\parallel NR.$$Now as $N,R$ are midpoints of $AQ$ and $BY'$ respectively, by an isoceles trapezoid's midline, we have that $QY'\parallel NR \parallel AB$. $\textbf{CLAIM: }Y'\equiv Y$. $\textbf{PROOF: }\measuredangle QY'C=\measuredangle QY'B\stackrel{QY'\parallel AB}{=}\measuredangle ABY'=\measuredangle ABC=\measuredangle ADC=\measuredangle QDC\implies DCY'Q$ is cyclic, and so we have $Y'\in\{BC, \odot(CDQ)\}\implies Y'=BC\cap \odot(CDQ)=Y$. $\textbf{CLAIM: }AECY$ is cyclic. $\textbf{PROOF: }$Note that for proving $AECY$ is cyclic, it is enough to show that $A,Y$ form a coaxial pencil with $\{\odot(CBX),\odot(CED)\}$. $$\dfrac{\operatorname{Pow}_{\odot CED}(A)}{\operatorname{Pow}_{\odot CBX}(A)}=\dfrac{AD\cdot AN}{AX\cdot AC}=\dfrac{\dfrac{AD\cdot AQ}{2}}{AX\cdot AC}=\dfrac{AX\cdot AC}{2AX\cdot AC}=\dfrac{1}{2}$$ $$\dfrac{\operatorname{Pow}_{\odot CED}(Y)}{\operatorname{Pow}_{\odot CBX}(Y)}=\dfrac{YC\cdot YR}{YC\cdot YB}=\dfrac{YR}{YB}=\dfrac{\dfrac{YB}{2}}{YB}=\dfrac{1}{2}$$ So, $\dfrac{\operatorname{Pow}_{\odot CED}(Y)}{\operatorname{Pow}_{\odot CBX}(Y)}=\dfrac{\operatorname{Pow}_{\odot CED}(A)}{\operatorname{Pow}_{\odot CBX}(A)}=\dfrac{1}{2}$ and we are done by Forgotten Coaxiality Lemma. So, from the claim above, we conclude that our redefined point $E$ is $=\odot(CBX)\cap \odot(CAY)$ indeed. $\textbf{CLAIM: }\measuredangle BCE=\measuredangle MDA$. $\textbf{PROOF: }\measuredangle BCE=\measuredangle BPE=\measuredangle MPN=\measuredangle MDN=\measuredangle MDA$. So, to finish notice that the value $\measuredangle MDA$ is fixed as $M$ is the midpoint of $AB$, and does not depend upon the position of $X$. So, we conclude that $\textbf{the locus of all such points}$ $E$ $\textbf{is the line through}$ $C$ $\textbf{such that}$ $\measuredangle BCE=\measuredangle MDA$.

Invert about $C$ with arbitrary radius. Let $P'$ denote the image of $P$ under this inversion. Let $(CAY)\cap (CBX)=Z$. So, $A'Y'\cap B'Y'=Z'$ and $A'B'\cap X'Y'=D'$. Let $A'B'\cap CZ'=Q'$ So, $Q$ is fixed point on $(ABCD)$ because using cevian-concurrency, $(A'B', Q'D')=-1\implies (AB, QD)=-1$. Thus, locus of $Z$ is line passing through $C$ and $Q$.

First note that by spiral similarity, triangle $DAB \sim DXY$, and if $M = (ACY) \cap (BCX)$ then $MXA \sim MBY$. Thus, $\frac{MX}{MB} = \frac{DA}{DB}$, so if $A’$ is the reflection of $A$ over $D$ then $BMX \sim BDA’$. Thus, $\angle{XCM} = \angle{A’BD}$ so $M$ ranges along the line through $C$ forming an angle equal to $\angle{A’BD}$ with $AC$.

Let $O$ and $O_1$ be the circumcenters of $ABC$ and $CXY$, $l$ be the perpendicular bisector of $CD$. Perpendicular from $O_1$ to $AC$ and perpendicular from $O$ to $BC$ intersect at $P$. Similarly define $Q$. Notice $P$ and $Q$ are circumcenters of $CBX$ and $CAY$ So the other intersection is reflection of $C$ across $PQ$.Let $Z$ be the infinity point of line $PQ$ First we say that $PQ$ moves parallel when $O_1$ moves across $l$. This can be seen easily because of the homothety with center $O$. So desired point lies on the perpendicular from $C$ to $PQ$ which doesn't change. Say that line $k$ Notice $-1=(OP,OQ;OO_1,OZ)=(CB,CA;CD,k)$ Where last equality holds because all corresponding lines are perpendicular which doesn't change the cross ratio between them. Finally $k$ is well-defined. Q.E.D.

Lemma : Let $ABC $ be a triangle and $M_1,M_2,M_3$ variable points on a line $d$ $P_1,P_2,P_3$ their projections on $CB$ and $Q_1,Q_2,Q_3$ their projections on $CA$ then $(ACP_i) \cap(BCQ_i)$ are colinear. proof Let $S_1 $ the similicenter of the similarity that sends $AQ_1\to P_1B$ then $S_1=(AP_1C)\cap (BQ_1C)=\{C,S_1\} $ and $SAP_1\sim SQ_1B$; Let $ S_2 $ the similicenter of the similarity that sends $AQ_2\to P_2B$ then $S_2=(AP_2C)\cap (BQ_2C)=\{C,S_2\} $ and $SAP_2\sim SQ_2B$; Let $ S_3 $ the similicenter of the similarity that sends $AQ_3\to P_2B$ then $S_3=(AP_3C)\cap (BQ_3C)=\{C,S_3\}$ and $SAP_3\sim SQ_3B$; we have $\frac{P_1P_2}{P_1P_3}=\frac{Q_1Q_2}{Q_1Q_3}$ then by the linked theorem1 we get $SAP_3\sim SQ_3B$ where $S$ is point of $S_1S_2$ s.t. $\frac{S_1S_2}{S_1S_3}$ hence $S_3=S$ therefore $ S_1,S_2,S_3$ are colinear . Back to the problem : Let the variable circle cuts $DB$ at $Z$ by Reim's we deduce that $ ZY\parallel AB,ZX\parallel BB$ ($BB$ is the tangent to $(ABC)$ at $B$) applying the lemma with $ABC$ and $d=BB$ the result foolows.

Wow, fantastic problem!!!

It has become so interesting that even mere pictures not even <configs are inverted for profit nowadays. R.I.P Geo.......

Idk probably over-overkill: First we notice that $\angle XDY = \angle XCY = \angle ACY$ which is constant, so assuming $X$ moves on $AC$, we can see that $X$ maps to $Y$ projectively. Inverting at $C$ with arbitrary radius gives us a figure where we have a projective map between the pencils of lines passing through $A$ and $B$ ($\odot BCX$ turns into a line $BX$ and similarly with $A$), and using Steiner conic theorem this means that there exist a line passing through all intersections of $BX$ and $AY$, checking $X=C$ gives us that $C$ lie on this line, so going back to the original figure the locus is indeed a line passing through $C$, and for more specification we could try more cases..

Let $ACY$ and $BCX$ meet at $K$. Let $\angle ACK = \alpha$ and $\angle BCK = \beta$. Note that $\alpha + \beta$ is fixed and $\frac{\sin{\alpha}}{\sin{\beta}} = \frac{KX}{KB}$ and since $KXA$ and $KBY$ are similar we have $\frac{KX}{KB} = \frac{AX}{BY}$ and since $DXA$ and $DYB$ are similar we have $\frac{AX}{BY}= \frac{DA}{DB}$ which is fixed. now since $\frac{\sin{\alpha}}{\sin{\beta}}$ is fixed we have that $K$ moves on a line passing through $C$. Let this line meet $AB$ at $S$ and let $AB,CD$ meet at $T$. Note that $\frac{AS}{SB} = \frac{AC}{CB}.\frac{\sin{\alpha}}{\sin{\beta}} = \frac{AC}{CB}.\frac{AD}{DB} = \frac{TA}{TB}$ so $K$ lies on line $CS$ such that $(TS,AB)=-1$.