Dear Bertrand & Ross - could you check the following? I neither speak French nor Math :-)
but I computed the same skein relations with magic and trickery (so my results
are unproven, of course). I get 11+6+2*8+2*3=39 basis 3+3 tangles (which I here
already split into the D6h symmetry classes) and under the E7 family polynome
10+5+2*7+2*3=35 are independent. The linear rest should be your skein relations.
So, are there exactly 4 (including symmetry-transformed!) of them, i.e. your Figs. 2/3?
Since they don't look symmetric under rotation, it's hard to check for me.

(BTW, for 2+2 tangles, I "know" an even generalized result (of your Fig. 1) for 20 years :-)

Maybe going to 4+4 tangles could solve the problem of having a complete
reducing skein set (or at least give an unproven solution) but since that might
need hundreds of basis tangles you a) either have to crowdsource the
calculation or b) do it with birdtracks or c) use math (I'm out then :-)

EDIT: Now that I'm back at my PC with access to the literature, 4+4 doesn't look
too promising either. Kuperbergs G2 paper says there are 455 crossingless freeways
with 8 endpoints. Uck. (And under E7 instead, a lot of them are inaccessible.)
So if one could do the following:
1. Prove that: IF you can reduce 1-gons and 2-gons and two adjacent 3-gons
(works for all the E7 family) AND an adjacent 3- and 4-gon pair AND two 3-gons
touching on a corner (five crossings; i.e. simpler diagrams now span the 8-endpoint
vector space...or so I hope!) THEN you could reduce any link diagram, and:
2. the compatibility rules of this reducing set are finite (I don't have much
hope for that either!)...
the whole E7 family polynomial proof would reduce to skein diagram manipulation
(of gargantuan size, of course).
Yup, it's probably better to use math instead :-)