In my notes I have the following claim.
Let $G$ be a simply connected abelian Lie group. Since $G$ is abelian, we have that $exp(A+B) = exp(A) exp(B)$ for all $A,B∈Lie(G)$. Therefore, the exponential map is an isomorphism between the additive group $Lie(G)$ and $G$.
I think I understand surjectivity because we know $exp : Lie(G) \to G$ restricts to a diffeomorphism on a neighbourhood $U\ni 0$. And since $G$ is connected, we have that $\langle U \rangle=G$ so any element $g\in G$ can be written $g=exp(X_1)^{\pm 1}\ldots exp(X_n)^{\pm 1}=exp(\pm X_1 \pm \ldots \pm X_n)$ (because $G$ is $G$ is abelian so $Lie(G)$ is abelian as well). But why is it injective and what does it has to do with simple connectivity?
The lie algebra $\mathfrak{g}$ is an abelian Lie group, and as you pointed out, $\exp:\mathfrak{g}\to G$ is a homomorphism. Since there is a neighbourhood of $0$ on which $\exp$ is a diffeomorphism, the kernel $\ker (\exp)$ of this homomorphism is discrete. Therefore $\exp$ is a covering map. Since $G$ is simply-connected, it must be a homeomorphism, and in particular bijective. So it is an isomorphism.
EDIT. Regarding the discreteness of the kernel:
Let $U\ni 0 $ be a neighbourhood of zero such that $\exp:U\to \exp(U)$ is a diffeomorphism. In particular, for $v\in U\setminus\{0\}%$, we have $\exp(v)\neq\exp(0)=1$, since $\exp$ is injective on $U$. So $K\cap U = \{0\}$. But $K\cap U$ is open in $K$ in the subspace topology by definition, so $\{0\}$ is open in $K$, which means $K$ is discrete.