I am trying to show that if $[X,Y]=0$ then the exponential map $\exp : Lie(G)\to G$ is such that $$\exp(X+Y)=\exp(X)\exp(Y), \forall X,Y\in Lie(G).$$ The hint is to show that $\gamma : t\mapsto \exp(tX)\exp(tY)$ is a one-parameter subgroup, which I can show : \begin{align} \gamma(t+s)=&\exp((t+s)X)\exp((t+s)Y)\\ =&\exp(tX)\exp(sX)\exp(tY)\exp(sY)\\ =&\exp(tX)\exp(tY)\exp(sX)\exp(sY)\\ =&\gamma(t)\gamma(s) \end{align}
and
\begin{align} (\gamma(t))^{-1}&=(\exp(tX)\exp(tY))^{-1}\\ &=\exp(tY)^{-1}\exp(tX)^{-1}\\ &=\exp(-tY)\exp(-tX)\\ &=\exp(-tX)\exp(-tY)\\ &=\gamma(-t) \end{align}
Note I used a result of a previous exercise that says that $[X,Y]=0$ iff $\exp(tX)\exp(sY)=\exp(sY)\exp(tX), \forall s,t \in \mathbb{R}$
But how to use that $\gamma$ is a one parameter subgroup to conclude?
Two Hints:
Observe that all the one-parameter subroup are linked with the exponential map (how?)
Recall that the differential of the group multiplication act like the sum, i.e.
$$ d_{(e,e)}(\cdot)((v, w)) = v + w $$
where we identify $T_{(e,e)}(G \times G) = T_e G \times T_e G$.
I will upgrade this hints to a full proof once you solved the problem