"Then the claim would be immediate by
semisimplicity if one can show that
every irreducible representation of
$G(\mathbb{C})$ (or perhaps of Lie
groups in some more general class than
these) occurs as a subquotient of a
tensor power of a faithful one. How
might one prove the latter? (Can one
prove the latter for compact real
groups in a manner similar to the
proof for finite groups, and then pass
to semisimple complex groups by the
unitary trick?)"
Yes. The analysis is not so pretty, but it is elementary. Let $K$ be a compact Lie group, $V$ a faithful representation, and $W$ any other representation. Just as in the finite group case, $\mathrm{Hom}_K(W,V^{\otimes N}) \cong (W^* \otimes V^{\otimes N})^K$, and the dimension of the latter is $\int_K \overline{\chi_W} \cdot \chi_V^{N}$, where $\chi_V$ and $\chi_W$ are the characters of $V$ and $W$, and the integral is with respect to Haar measure. Let $d_V$ and $d_W$ be the dimensions of $V$ and $W$.
We now come to a technical nuisance. Let $Z$ be those elements of $K$ which are diagonal scalars in their action on $V$; this is a closed subgroup of $S^1$. For $g$ not in $Z$, we have $|\chi_V(g)| < d_V$. We first present the proof in the setting that $Z = \{ e \}$.
Choose a neighborhood $U$ of $\{ e \}$ small enough to be identified with an open disc in $\mathbb{R}^{\dim K}$. On $U$, we have the Taylor expansion $\chi_V(g) = d_V \exp(- Q(g-e) + O(g-e)^3)$, where $Q$ is a positive definite quadratic form; we also have $\chi_W(g) = d_W + O(g-e)$. Manipulating $\int_U \overline{\chi_W} \chi_V^N$ should give you
$$\frac{d_W \pi^{\dim K/2}}{\det Q} \cdot d_V^N \cdot N^{-\dim K/2}(1+O(N^{-1/2}))$$
Meanwhile, there is some $D<d_V$ such that $|\chi_V(g)| < D$ for $g \in K \setminus U$.
So the integral of $\overline{\chi_W} \chi_V^N$ over $K \setminus U$ is $O(D^N)$, which is dominated by the $d_V^N$ term in the $U$ integral.
We deduce that, unless $d_W=0$, we have $\mathrm{Hom}_K(W, V^{\otimes N})$ nonzero for $N$ sufficiently large.
If $Z$ is greater than $\{e \}$, then we can decompose $W$ into $Z$-isotypic pieces. Let $\tau$ be the identity character of the scalar diagonal matrices, and let the action of $Z$ on $W$ be by $\tau^k$. (If $\tau$ is finite, then $k$ is only defined modulo $|Z|$; just fix some choice of $k$). Then we want to consider maps from $W$ to $V_N:=V^{k+Nd_V} (\det \ )^{-N}$. $V_N$ is constructed so that $\overline{\chi_W} \chi_{V_N}$ is identically $d_W$ on $Z$; one then uses the above argument with a neighborhood of $Z$ replacing a neighborhood of the origin.
A solvable reductive group has a very simple structure - the component group is solvable, and the connected component of the identity is an algebraic torus. Unfortunately it is not really possible to decompose it further, because the conjugation action would need to decompose as well, but there are many indecomposable actions of solvable finite groups on tori. These basically boil down to irreducible solvable subgroups of $GL_n(\mathbb Z)$, $n$ the dimension of the torus. There are lots of these.
For nilpotent groups, the situation is a bit better. The conjugation action must be trivial, so you can write the group as the almost direct product of a central torus and a finite group as follows: The torus is the connected component of the identity. The finite group is the kernel of the universal map from the group to a torus.
Best Answer
Question $1$: The theorem of Mostow says that every connected algebraic group $G$ over a field $K$ of characteristic zero has a Levi decomposition. This means, $G$ has a reductive algebraic subgroup $S$ such that $G$ is the semidirect prodcut of $S$ and $R_u(G)$, the unipotent radical of $G$. Moreover, any reductive algebraic subgroup $S'$ of $G$ is conjugate to $S$. The result is not true in general in prime characteristic. There is a nice article by Jim Humphreys, EXISTENCE OF LEVI FACTORS IN CERTAIN ALGEBRAIC GROUPS, in the pacific journal of mathematics $1967$.
Question $2$: The Levi decomposition for Lie groups was shown by Levi and Malcev.