Well Lie algebras naturally arise from the Lie bracket of vector fields and from taking the Lie algebra of a Lie group. If we look at a the Lie algebra of a matrix subgroup, then the Lie bracket is the commutator of matrices.
What kind of ring is $U(L)$?
Since representations of Lie algebras behave like representations of groups (the category has tensor products and duals, for example), you should expect that the universal enveloping algebra $U(\mathfrak{g})$ has some extra structure which causes this, and it does: namely, it is a Hopf algebra (a structure shared by group algebras). The comultiplication is defined on basis elements $x \in \mathfrak{g}$ by
$$x \mapsto 1 \otimes x + x \otimes 1$$
(this is necessary for it to exponentiate to the usual comultiplication $g \mapsto g \otimes g$ on group algebras) and the antipode is defined by
$$x \mapsto -x$$
(again necessary to exponentiate to the usual antipode $g \mapsto g^{-1}$ on group algebras).
This is an important observation in the theory of quantum groups, among other things.
Thus, via the envelopping algebra Lie algebras and their represnetations cn be studied from a ring theoretic point of view. Is the cconverse true in some sense?
Not in the naive sense, the basic problem being that if $A$ is an algebra and $L(A)$ that same algebra regarded as a Lie algebra under the bracket $[a, b] = ab - ba$, then a representation of $L(A)$ does not in general extend to a representation of $A$, but to a representation of $U(L(A))$, which may be a very different algebra (take for example $A = \text{End}(\mathbb{C}^2)$).
Of course there are other relationships between ring theory and Lie theory. For example, if $A$ is a $k$-algebra then $\text{Der}_k(A)$, the space of $k$-linear derivations of $A$, naturally forms a Lie algebra under the commutator bracket. Roughly speaking this is the "Lie algebra of $\text{Aut}(A)$" in a way that is made precise for example in this blog post.
Best Answer
Every associative algebra $A$ has a canonical Poisson algebra structure where the bracket is given by commutator $\{ a, b \} = ab - ba$, as you say. This is not the only possible Poisson bracket on $A$. This particular bracket vanishes iff $A$ is commutative, but commutative algebras can have other interesting (and in particular nonvanishing) Poisson brackets on them (and so can associative algebras).
For example, $\mathbb{R}[x, y]$ has a Poisson bracket satisfying $\{ x, y \} = 1$ (all other brackets are determined by the Leibniz rule); this is the Poisson algebra of polynomial functions on the cotangent bundle $T^{\ast}(\mathbb{R})$, which is even a symplectic manifold. Also, if $\mathfrak{g}$ is any Lie algebra, the symmetric algebra $S(\mathfrak{g})$ has a Poisson bracket extending the Lie bracket on $\mathfrak{g}$; this is the Poisson algebra of polynomial functions on the Poisson manifold $\mathfrak{g}^{\ast}$.