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.
Your question whether $\mathfrak{g}$ is semisimple is equivalent to whether necessarily $\mathfrak{r} = 0$. The answer is no. A counterexample is given by
$$\mathfrak{g} = \left\{ \pmatrix{a & b & d\\
c& -a& e\\
0&0&0} : a,b,c,d,e \in \Bbb C\right\}$$
where the radical
$$\mathfrak{r} = \left\{ \pmatrix{0 & 0 & d\\
0& 0& e\\
0&0&0} : d,e \in \Bbb C\right\}$$
is two-dimensional.
Best Answer
In each name, the word "semisimple" means "a direct sum of simple objects" in the appropriate sense. It can be shown that a complex Lie algebra is semisimple (has radical zero) if and only if it is a direct sum of simple Lie algebras. On the other hand, saying that $\operatorname{ad} x$ is diagonalizable is the same as saying that $\mathfrak{g}$ decomposes into a direct sum of eigenspaces for $\operatorname{ad} x$, on each of which $\operatorname{ad} x$ acts as a scalar multiple of the identity; in other words, $\operatorname{ad} x$ is a "direct sum of scalars".