Let $G$ be a Lie group, and $\xi \in T_{e}G$ a tangent vector at the identity. Given a function $f \in C^{\infty}(G)$, verify that $ g \rightarrow ((\ell_{g})_{*}\xi)f$ is a $C^{\infty}$ function on $G$, and deduce that $X_{\xi}: f \rightarrow (g \rightarrow ((\ell_{g})_{*}\xi)f)$ defines a vector field on $G$.
Ultimately, all this is saying is that the Lie algebra $\mathfrak{g}$, i.e. the set of left-invariant vector fields on $G$, defines a natural linear map $\mathfrak{g} \rightarrow T_{e}G$, for which we have $\mathfrak{g} \ni X \rightarrow X_{e} \in T_{e}G$ such that this is a linear isomorphism.
Proving injectivity is easy. We just take any $g \in G$. Then $X_{g}=\ell_{g*}X_{e}$, so our map is injective. Now we need only prove surjectivity.
First, is my interpretation of the problem correct? Second, how would one show surjectivity? Thanks very much!
Curiously your proof of injectivity
also provides the proof of surjectivity. I think we need to be a bit more verbose to see the difference between the two uses of this statement.
Injectivity: Suppose that $X$ and $Y$ are left invariant vector fields with $X_e = Y_e$. In order to show that $X = Y$ we just take any g∈G. Then $X_g=ℓ_{g^∗}X_e$ and similarly $Y_g=ℓ_{g^∗}Y_e = ℓ_{g^∗}X_e$, hence $X_g = Y_g$.
Surjectivity: Suppose that $\xi \in T_eG$. We want to show that there exists a left invariant vector field $X$ with $X_e = \xi$. We just take any g∈G. Then $X_g=ℓ_{g^∗}X_e = ℓ_{g^∗}\xi$ defines the desired $X$. (Of course this is just the same thing you say at the beginning of your post.)
Actually what I think causes the confusion is that in the text at the beginning of your post you construct vector field out of a tangent vector and then you interpret this construction as a map going the other way (i.e. making a tangent vector out of a vector field). So instead of showing injectivity and surjectivity you might also consider checking that the two constructions are indeed eachother's inverses.