Consider a one-sided ( say, internal) neighborhood $U$ of the unit circle $S$ ( $U$ contains $S$) on the plane with a choice of smooth complex stricture $\tau$ on $U$.


By smoothness of $\tau$ on $U$ we mean that corresponding operator of almost-complex structure $J$( defined in the interior of $U$) as a matrix with respect to the coordinate basis on the plane, has smooth extension to $S$.


Two such pairs $(U\_1,\tau\_1)$ and $(U\_2,\tau\_2)$ are equivalent if there are subneighborhoods $V\_1\subset U\_1$ and $V\_2\subset U\_2$ ( both containing $S$) and a diffeomorphism $f: V\_1\rightarrow V\_2$ which is biholomorphic in the interior of $V\_1$ with respect to $\tau\_1, \tau\_2$.


What is the moduli space of such structures?



What is the moduli space of germs of one-sided complex structures near the circle?
I think that, even if you include the boundary $S$, they are all equivalent. In saying so, I'm assuming that the final 'it' in your first sentence refers to the plane, not to $U$ and that you are considering smooth complex structures. 


The reason is this: Take a sufficiently small open neighborhood $W$ of the closure of the disk $\Delta$ that $S$ bounds that is also a disk. By the Riemann mapping theorem, there is a biholomorphism of $W$ (endowed with the given complex structure) with the unit disk $\Delta$ endowed with its standard complex structure. 


The image $S'\subset \Delta$ of $S$ under that biholomorphism is a smoothly embedded curve that bounds a disk $D'\subset\Delta$. By the known boundary regularity in the Riemann mapping theorem, there is a map $(D',S')\to (\Delta,S)$ that is a biholomorphism in the interior and smooth up to and including the boundary (see the discussion at [Riemann mapping theorem and smoothness on the boundary](https://mathoverflow.net/questions/82613)). Composing these two biholomorphisms gives the equivalence of the original $(U,S)$ with the given complex structure with a $(U',S)$ for which the complex structure is the standard one.


Now pass to germs, and you have the equivalence that you wanted.


