There's a very neat argument to see the bijection between representations of a group $G$ on $k$-vector spaces $V$ and $kG$-modules $V$: There exists an adjunction $$U: k-alg \leftrightarrow Grp: A $$ $$A \mapsto A^{\times} $$ $$kG \mathrel{\unicode{x21a4}} G $$ where $A^{\times}$ is the group of multiplicative units of $A$. This gives a bijection $$\hom_{Grp}(G,GL(V)) = \hom_{Grp}(G,U(End(V))) \cong \hom_{k-alg}(A(G),End(V)) = \hom_{k-alg}(kG,End(V)) $$ which is exactly what we wanted! A similar argument applies to a Lie algebra $\mathfrak{g}$ and modules over its universal enveloping algebra $U(\mathfrak{g})$.
However, to use this in practice, we also want to make sure that this bijection respects maps of representations. One can easily do this "by hand", but it would be really nice if there were an entirely categorical argument to show this.
Does there exist a categorical argument to show the equivalence of categories $Rep(G)\cong Rep(kG)$?
Let me be a bit more specific about what "entirely categorical" means. (Here we will consider monoid representations rather than group representations but the difference is minimal).
Let $\mathcal{C},\mathcal{D}$ be categories, $a: ob(\mathcal{C}) \to ob(\mathcal{D})$ be a function and $F: \mathcal{D} \to \mathcal{C}$ be a functor. Suppose that $F(a(x)) = \hom(x,x)$. For $y\in\mathcal{D}$, a representation of $y$ in $\mathcal{C}$ is a choice of $x\in\mathcal{C}$ and $\mathcal{D}$-morphism $f: y \to a(x)$. Given two $y$-reps, a morphism of representations $(x_1,f_1)\to(x_2,f_2)$ is a $\mathcal{C}$-morphism $\varphi: x_1\to x_2$ such that $$\hom(x_1,\varphi)\circ F(f_1) = \hom(\varphi,x_2)\circ F(f_2) $$ This defines the category $Rep(y)$.
Now let $(\mathcal{C}_1,\mathcal{D}_1,a_1,F_1,y_1)$ and $(\mathcal{C}_2,\mathcal{D}_2,a_2,F_2,y_2)$ be two such collections of data and suppose there exists an adjunction
$$A: \mathcal{D}_1 \leftrightarrow \mathcal{D}_2: B$$
such that $y_2 = A(y_1)$ and $a_1(x) = B(a_2(x))$. Let $\psi_{t,s}: \hom(t,B(s))\cong \hom(A(t),s)$ be the induced hom-set bijection. There is an obvious map
$$H: Rep(y_1) \to Rep(y_2)$$ $$(x,f) \mapsto (x,\psi_{y_1,a_1(x)}(f)) \text{ on objects}$$ $$\varphi \mapsto \varphi \text{ on morphisms}$$
which is clearly an isomorphism of categories, so long as it is a functor.
Is $H$ a functor? That is, does it really take morphisms to morphisms?
fosco's argument works but, as you say, doesn't generalize in an obvious way to the Lie algebra case. Here's a different argument that does (or at least a sketch).
In general, suppose $C$ is a cocomplete $k$-linear category equipped with a forgetful functor $U : C \to \text{Vect}$. We would like to know: when does $U$ exhibit $C$ as the category $\text{Mod}(A)$ of modules over a $k$-algebra $A$ (in such a way that $U$ is the usual forgetful functor)? For starters, a necessary condition is that $U$ must be a representable functor, with representing object $P$ satisfying $\text{End}(P) \cong A$. What else is necessary?
The answer is that it's necessary and sufficient that $P$ be a compact projective generator, meaning that
This is Gabriel's theorem. To apply it to $\text{Rep}(G)$ (or more generally representations of a monoid) we need to check that the forgetful functor preserves all colimits (which follows from the general fact that colimits in a functor category are computed pointwise, and that $\text{Rep}(G)$ is the category of functors $BG \to \text{Vect}$) and is faithful (pretty much by definition), and is representable.
Representability is the interesting condition. I believe it follows abstractly from the adjoint functor theorem (or, more or less equivalently, the representable functor theorem), but it's a bit annoying to check the solution set condition. We can instead directly verify that $k[G]$ with the usual left $G$-action represents the forgetful functor $\text{Rep}(G) \to \text{Vect}$, or if we want to be a bit more categorical about it, we can argue as follows:
The representing object must in fact represent the underlying set functor $\text{Rep}(G) \to \text{Vect} \to \text{Set}$, and the underlying set functor factors in a different way as the composite $\text{Rep}(G) \to G\text{-Set} \to \text{Set}$. Both of these functors have left adjoints, namely the free $G$-set functor $X \mapsto G \times X$ and the free vector space functor $X \mapsto k[X]$. The composite of these left adjoints is the left adjoint of the composite, so we conclude that the left adjoint of the underlying set functor $\text{Rep}(G) \to \text{Set}$ is the functor $X \mapsto k[G \times X]$ and in particular that it is represented by $k[G]$ with the usual $G$-action, which has endomorphism algebra $k[G]$ as desired.
Now for the Lie algebra case. Here we again need to check that the forgetful functor $\text{Rep}(\mathfrak{g}) \to \text{Vect}$ preserves colimits (this bit seems a little tedious, I can't think of a really clean argument for this), is faithful (again pretty much by definition), and is representable.
Again representability is the interesting condition. Again I believe it follows abstractly from the adjoint functor theorem but again it's better just to construct the representing object. I can't think of a really clean argument for this either but you can do it "by hand." The representing object $P$ of the forgetful functor $\text{Rep}(\mathfrak{g}) \to \text{Vect}$ must have a special vector $1 \in P$ such that evaluation at $1$ produces the identification $\text{Hom}_{\mathfrak{g}}(P, V) \cong V$. Furthermore since $P$ is a representation of $\mathfrak{g}$ we need to be able to apply every element of $\mathfrak{g}$ to this special vector; this produces a copy of $\mathfrak{g}$ in $P$. Repeatedly multiplying by elements of $\mathfrak{g}$ we get copies of all the tensor powers of $\mathfrak{g}$, so overall we construct a natural map $T(\mathfrak{g}) \to P$ from the tensor algebra of $\mathfrak{g}$ to $P$, and then working through what the kernel of this map must be we get the usual universal enveloping algebra relations, producing an isomorphism $P \cong U(\mathfrak{g})$ where $U(\mathfrak{g})$ has the usual $\mathfrak{g}$-action, and endomorphism algebra $U(\mathfrak{g})$, as desired.