The correspondence holds for complex Lie groups. I'm not sure if the proof of the real case can be reproduced, but we can instead use the real case in our favor.
Given a complex simply connected Lie group $G$ of complex dimension $n$, we can regard $G$ as a real Lie group of real dimension $2 n$. The main idea is to use the fact that we can obtain the real Lie algebra of $G$ -- the algebra of smooth left invariant fields -- by restricting the scalars of the complex Lie algebra of $G$ -- the algebra of holomorphic left invariant vector. This is because every invariant vector field is holomorphic.
Since $G$ is simply connected, the map $\operatorname{Hom}_{\mathbb{R}}(G, \operatorname{GL}_n(\mathbb{C})) \to \operatorname{Hom}_{\mathbb{R}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$ that takes a smooth group homomorphism $\varphi : G \to \operatorname{GL}_n(\mathbb{C})$ to the real Lie algebra homomorphism $\varphi_* : \mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$ is a bijection. Now consider the subset $\operatorname{Hom}_\mathbb{C}(G, \operatorname{GL}_n(\mathbb{C})) \subset \operatorname{Hom}_{\mathbb{R}}(G, \operatorname{GL}_n(\mathbb{C}))$ of holomorphic group homomorphisms. We will show that our map takes $\operatorname{Hom}_\mathbb{C}(G, \operatorname{GL}_n(\mathbb{C}))$ to the subset $\operatorname{Hom}_\mathbb{C}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C})) \subset \operatorname{Hom}_{\mathbb{R}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$ of complex Lie algebras homomorphisms.
Given a smooth homomorphism $\varphi : G \to \operatorname{GL}_n(\mathbb{C})$, if $\varphi$ is holomorphic then its derivative $\varphi_*$ is $\mathbb{C}$-linear, so it lies in $\operatorname{Hom}_{\mathbb{C}}(\mathfrak{g}, \mathfrak{gl}_n(\mathbb{C}))$. On the other hand, if $\varphi_*$ is $\mathbb{C}$-linear then $\varphi$ is holomorphic, so that $\varphi$ lies in $\operatorname{Hom}_\mathbb{C}(G, \operatorname{GL}_n(\mathbb{C}))$.
We should note that this is a particular case of the fact of a more general theorem which states that the map $\operatorname{Hom}_{\mathbb{C}}(G, H)) \to \operatorname{Hom}_{\mathbb{C}}(\mathfrak{g}, \mathfrak{h})$ is well defined and is a bijection for all complex simply connected $G$ and all complex $H$. Indeed, our proof still works if we replace $\operatorname{GL}_n(\mathbb{C})$ with any $H$. I hope I'm not missing any detail in here, but I was convinced by this proof.
Here is the answer in the form of a two-part exercise.
Exercise 1a: Let $G$ be a connected Lie group, $\rho : G \to GL(V)$ be a finite-dimensional representation of $G$ over $\mathbb{R}$ or $\mathbb{C}$, and $d \rho : \mathfrak{g} \to \mathfrak{gl}(V)$ be its derivative. Let $W \subseteq V$ be a subspace of $V$. Then $W$ is invariant under $G$ iff it's invariant under $\mathfrak{g}$.
Exercise 1b: Deduce that if $V$ is completely reducible (e.g. if $G$ is compact) then the irreducible decomposition of $V$ as a representation of $G$ coincides with its irreducible decomposition as a representation of $\mathfrak{g}$.
Best Answer
The short answer: Because $GL(V)$ is a group and $\mathrm{End}(V)$ is a Lie algebra.
In particular, $GL(V)$ has the following:
Whereas $\mathrm{End}(V)$ has the following:
So it is only natural to represent a Lie group in the group $GL(V)$, and a Lie algebra in the algebra $\mathrm{End}(V)$.