I initially wanted to prove that if a Lie group $G$ is abelian, then $\exp(X+Y) = \exp(X)\exp(Y)$ for any $X, Y \in \frak{g}$. I showed that: $$ \frac{d}{dt}(\exp(tX)\exp(tY))= d_e (l_{\exp(tX)\exp(tY)}) (Ad(\exp(-tY))(X)) + d_e l_{\exp(tX)\exp(tY)}(Y) $$ or at least I hope I have, if my calculations are correct. If this formula is correct, then obviously if $G$ is abelian then $Ad$ is trivial and $\exp(tX)\exp(tY)$ is an integral curve of $X+Y$ as wanted.
But is it true for any Lie group that $\exp(X+Y)=\exp(X)\exp(Y)$ if and only if $[X,Y] = 0$ (this question suggests it should be true also for Lie groups)? The formula above seems to get pretty close to it, but I don't know how to make progress from there.
I know also that in general on a manifold $[X,Y] = 0$ if and only if the flows commute, which in this case just means $\exp(tX)\exp(sY) = \exp(sY)\exp(tX)$, but I don't know if this helps.
Essentially, is it true that $Ad(\exp(-tY)(X)) = X$ for any $t$ if and only if $[X,Y]=0$?
The condition $[X,Y]=0$ does indeed imply that $\exp(X) \exp(Y) = \exp(X+Y)$.
This can be proved by applying the Baker-Cambell-Hausdorff formula which gives a Lie-algebra formula for a vector $Z$, expressed as a formal infinite series in terms of iterated commutators, that satisfies the equation $$\exp(X) \exp(Y) = \exp(Z) $$ (one may need to assume that $X,Y$ are "sufficiently small" in order to guarantee convergence of the formal series).
It's quite a complicated formula. However, in the special case that the 1st order commutator $[X,Y]$ satisfies $[X,Y]=0$, all the higher iterated commutators are also $0$ and the entire formula collapses to $Z=X+Y$.