I am trying to understand the connection between commutativity of a Lie group $G$ to the commutativity of its Lie algebra $\mathfrak{g}$.
A proof I am reading seems to use the fact that $X \in Z(\mathfrak{g})$ implies $e^X \in Z(G)$. I want to understand why is this correct.
I noticed that for any $x \in G$, $ad(x) = 0$ implies $x \in Z(g)$. Indeed, we can define $\gamma(t) = xe^{tY}x^{-1}$, and then $$\gamma'(t) = d(c_x)_{e^{tY}} Y_{e^{tY}} = d(c_x \circ L_{e^{tY}})_e Y_e = d(L_{xe^{tY}x^{-1}} \circ c_x)_e Y_e = d(L_{xe^{tY}x^{-1}})_e ad(x)(Y_e) = d(L_{\gamma(t)})_e Y_e = Y_{\gamma(t)}$$
(Where $c_x$ is the conjugation by $x$ map), showing that $\gamma(t) = e^{tY}$, so that $x$ commutes with $e^Y$ for all $Y$, and these generate $G$.
However, I don't see a good way to compute $ad(e^X)$, I only know that $ad(e^{tX}) \to ad(X)=0$ as $t\to 0$.
What you want to prove is the stronger fact that $X \in Z(\mathfrak{g})$ implies $\exp(tX) \in Z(G)$ for all $t \in \mathbb{R}$ (the converse is clear). Then it's clear what to do: $\text{ad}(\exp(tX))$ is the one-parameter subgroup of automorphisms of $\mathfrak{g}$ generated by $\text{ad}(X)$, and since the latter is zero by hypothesis, the former is the identity.
(Note that your reduction to considering $\text{ad}$ requires that $G$ be connected; this result is false otherwise.)