Proposition 2
Tag 0A9A, permanent. Cite it as [Bourbaki, Tag 0A9A].
Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra
The algebra \mathscr{T}^{(\infty)}(G) is associative. It is commutative if and only if G is commutative.
Let t \in \mathscr{T}^{(\infty)}(G), t' \in \mathscr{T}^{(\infty)}(G), t'' \in \mathscr{T}^{(\infty)}(G). Then t \* (t' \* t'') is the image of t \otimes t' \otimes t'' under the mapping (g, g', g'') \mapsto g(g'g'') of G \times G \times G into G and (t \* t') \* t'' is the image of t \otimes t' \otimes t'' under the mapping (g, g', g'') \mapsto (gg')g'' of G \times G \times G into G. Hence (t \* t') \* t'' = t \* (t' \* t''). It is seen similarly that, if G is commutative, t \* t' = t' \* t. If the convolution product is commutative, G is commutative by Proposition 1 (iii).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.