A straightforward approach for finding tilings of congruent, six-fold symmetrical tiles is to use a wallpaper-symmetrical tiling (I haven’t found any tilings that don’t have p6 or p6m symmetry). A trivial solution is to tile the plane with hexagons; if the symmetry axis of the tile doesn’t line up with the translation of the tiling, the solutions are more complex.
I have made a JS applet for drawing and generating these tilings, the page also has some example images https://istscilveks.itch.io/snowflake-tilings
It seems like all the snowflakes are either made of smaller hexagons or are fractals, depending on the angle. I suspect that the boundary can be described using a formal language, as has been done with double square polyomino tiles ("polyominoes yielding two distinct periodic tilings by translated copies such that every polyomino in the tiling is surrounded by exactly four copies") in Combinatorial properties of double square tiles. In fig. 11 some of the tiles are solutions to a related problem: tiling the plane with congruent, four-fold symmetrical tiles.
That paper and others on polyomino tilings, plus a few disparate solutions, are about all that I could find, so I would like to know if anyone has any resources or has heard of a problem like this.

I couldn't find any discussion of this problem online, so I went ahead myself and made an algorithm which can exhaustively generate the figures described, given a few extra assumptions. The linked paper turned out to be exactly what was needed to solve the problem, and here I give some of the steps for my modified algorithm (I haven't written the full algorithm in English). I defined a snowflake as a figure that has has all the symmetries of a regular hexagon, and a tileable snowflake as a snowflake for which the following assumptions also hold:
The assumptions allow the boundary of each tileable snowflake to be described using a combinatorial word: a list of digits that increase or decrease by one (modulo $6$), where each digit represents a unit step in a certain direction.
I also modified the operators from the paper and proved that:
Then, by Fermat’s infinite descent principle, these operators are sufficient to generate any tileable snowflake that conforms to my assumptions.
Additionally, I generalized the proof to other tilings: for any $p,q>3$, a snowflake is a figure with all the symmetries of a regular $p$-gon and a tileable snowflake admits a tiling that is topologically equivalent to the tiling with Schläfli symbol $\{p, q\}$. The center of each snowflake and each vertex of the tiling is assumed to be a center of symmetry of orders $p$ and $q$ respectively.
I also improved the app to allow generating snowflakes by inputting a string of operators.