Yes.

At least, rationally.

The result that you want starts on p68 of "Loop Groups" by Pressley and Segal. There, they prove surjectivity of H^{*}(L𝔤;ℝ) → H^{*}(LG;ℝ). The basic idea of the argument is as follows: for reasonably simple reasons, the cohomology of LG is easily obtainable from that of G. This yields specific formulae for generators of the de Rham cohomology of LG. Although these forms are not themselves left invariant, they are cohomologous to rational multiples of left invariant forms, and thus come from the cohomology of the Lie algebra, L𝔤. The proof of surjectivity is thus not hard.

The proof that it is an isomorphism is a little trickier and they defer that to section 14.6 (p299). That this is quite some way through the book is a good indication of how much you need to absorb to understand it.

Amazingly, parts of Loop Groups (including pages 68, 69, and 299, but not 300) are available on Google books. It says that it is a "Limited preview" but whether or not that is just limited in pages or limited in time, I do not know. However, Loop Groups is a great book for anyone interested in Lie Groups and infinite dimensional "stuff".

(Incidentally, this is all for G simply connected.)