It is rather immediate that the linear map $X_{e} \rightarrow (X : p \rightarrow dL_{p}(X_{e}))$ is a isomorphism of vectors paces when $p$ is a point of a left invariant Lie group.
However, I fail to understand why the Lie bracket on the the tanget space of $G$ at the identity is identical to the structure on the space of left invariant vectors fields of $G$. It should follow from the way we define it i.e $[X_{e},Y_{e}]=[X,Y]_{e}$, but I do not see why.
With which definitions do you want to work? We can define the Lie algebra $\mathfrak{g}$ associated to $G$ as left invariant vector fields on (connected component of) $G$, then $\mathfrak{g}$ has a bracket, induced by one on the vector fields. Equivalently, such vector fields are in bijection with tangent vectors at any given point on $G$, we choose $e$ just because we can (as a manifold, Lie group has the same geometry at any point and shifts allow us to use it: create a left invariant Haar measure, make symplectic form, prove orientability, etc).
You may wonder why the indentity $[X_{e},Y_{e}]=[X,Y]_{e}$ holds, it may seem that $X_{e}$ and $Y_{e}$ is something local, but $X_{e}, Y_{e}$ define left invariant fields $X, Y$ uniquely everywhere, then we take their bracket: $[X, Y]$ - this is again a left invariant field, which correspond to $[X, Y]_{e}$ in our tangent space $T_eG$, but it also correspond to a bracket $[X_e, Y_e]$ in $T_eG$ and thus they are equal.