I found this very nice theorem in the book by Onishchik ,"Lie groups and Lie algebras I".
Proposition 2.9. Let $g(t,s)\in G$ be a smooth map from some domain in $\mathbb{R}^2$ into a Lie group $G$. Define the corresponding maps into the Lie algebra $\mathfrak{g}$ of $G$ by $$ \xi(t,s)=\left(R_{g}^{-1}\right)_\ast\partial_t g,\quad \eta(t,s)=\left(R_{g}^{-1}\right)_\ast\partial_s g. $$ Then
$$ \partial_t \eta - \partial_s\xi=[\xi,\eta].$$
I'm familiar with this kind of formula from the theory of integrable systems, where it's called a "zero curvature condition". Obviously it also follows from the Maurer-Cartan equation by pulling back along $g$. Anyway, the proof of this is elementary in matrix groups because you just differentiate the two equations defining $\xi$ and $\eta$ with respect to the other variable and then notice that $\partial_t\partial_sg=\partial_s\partial_tg$. This is basically what the book does by using local coordinates.
What I'm struggling with is writing down a coordinate-free proof that doesn't rely on invariant vector fields, Maurer-Cartan forms, the exponential map, or any other high-level machinery. The equations make total sense just if you know what the tangent spaces $T_e G$ and $T_g G$ are, so it feels like one should be able to prove them through elementary calculus on manifolds. The book uses this equation to give an elementary proof of one of Lie's fundamental theorems without reference to the exponential map, which seems awfully nice. Does anyone know of a cleaner proof of this theorem? Or does it secretly rely on the exponential map somehow?
This might not be what you want, but I don't think you can prove the Maurer-Cartan equation without referencing differential forms, because the main object in the Maurer Cartan equation is the Maurer Cartan form, which is itself a differential form. What I can do is provide a coordinate free proof of the Maurer-Cartan equation.
Define the Maurer Cartan form on a Lie group $G$ pointwise by: \begin{align} \mu_G(X_g)=D_gL_{g^{-1}}(X_g) \end{align} and let $M\times G$ be the trivial principal bundle, with trivial connection/horizontal distribution $H=\pi_M^*TM$. Then clearly the connection one form inducing this connection is given by $\pi_G^*\mu_G$. The curvature vanishes identically as: $$F(X,Y)=d\mu_G(\pi_{G*}\circ \pi_{H} X,\pi_{G*}\circ\pi_{H}X)=d\mu_G(0,0)=0$$ since $\pi_{G*}\circ \pi_{H}$ is identically zero, where $\pi_H$ is the projection $T(M\times G)\rightarrow \pi_M^*TM$.
By the structure equation (which you can prove in a coordinate invariant way easily), we have that: \begin{align} d\pi_G^*\mu_G+\frac{1}{2}\left[\pi_G^*\mu_G,\pi_G^*\mu_G\right]=0 \end{align} Since $\pi_G^*$ commutes with the bracket and exterior derivative, we have that: \begin{align} \pi_G^*\left(d\mu_G+\frac{1}{2}\left[\mu_G,\mu_G\right]\right)=0 \end{align} Since the pull back by a submersion is injective, it follows that: $$d\mu_G+\frac{1}{2}\left[\mu_G,\mu_G\right]=0$$ thus we have the Maurer-Cartan equation.