Let $G$ be an abelian Lie group. The connected component $G_0$ is a connected abelian Lie group, so it's isomorphic to $\Bbb{R}^n\times\Bbb{T}^m$ for some $n,m\in\omega$ (this is not hard to see: mostly one just needs to observe $\exp$ is a surjective open homomorphism with a discrete kernel). In particular, it is divisible.
Lemma. Suppose $D$ is a divisible subgroup of an abelian group $G$ and $h:D\to H$ is a homomorphism into a divisible abelian group $H$. Then $f$ extends to a homomorphism $\tilde f:G\to H$.
For a proof, see here.
Therefore the identity $\mathrm{id}:G_0\to G_0$ extends to a homomorphism $f:G\to G_0$. Denote $K=\ker(f)$; clearly $K\cdot G_0=G$ and $K\cap G_0={0}$ so (since $G$ is abelian) $G=K\times G_0$. Since $G_0$ is open, we moreover know that $K$ must be discrete (and therefore closed). This shows $G=K\times G_0$ as a topological group and not just an abstract group.
Thus, we have seen abelian Lie groups are exactly the direct products of an abelian discrete group and a connected abelian Lie group.
If we moreover require $G$ to be compact, then $K$ must be finite (since it is compact, as a closed subgroup of $G$, and discrete) and $G_0$ a torus (since all connected compact abelian Lie groups are tori). So a group $G$ is a compact abelian Lie group if and only if it is isomorphic to $F\times \Bbb{T}^m$ for a finite abelian group $F$ and $m\in\omega$.
The Lie algebra, $\frak{g}$, of a compact Lie group, $G$, is reductive, meaning its a direct sum of an abelian Lie algebra and a semisimple Lie algebra. This is not difficult to prove. Here's a proof
http://books.google.com/books?id=PAJmnm3DU1QC&pg=RA1-PA249#v=onepage&q&f=false
(The proof there uses an equivalent definition of reductive: every ideal has a complimentary ideal. But proving the equivalence is not difficult.)
If $\frak{g}$ is solvable, the semisimple part must be zero. So $\frak{g}$ is abelian. Therefore $G$ is a torus.
If I've skipping too many details, please let me know.
Best Answer
The hardest part of the problem is showing:
If $G$ is a connected abelian Lie group then the exponential map $$ \exp: \mathfrak{g} \longrightarrow G $$ is a $\textbf{surjective homomorphism}$ with $\textbf{discrete kernel}$.
This can be done by appealing to the fact that $G$ is generated by the image of $\exp$, the first Lie theorem that states $\exp$ is a local homeomorphism at the identity, 1 parameter subgroups, and general considerations on topological groups - it's a good exercise to see if you've grasped the theory.
Our result then follows by noting that $G \simeq \mathbb{R}^k / \Gamma$ where $k$ is the dimension of $\mathfrak{g}$ as a real vector subspace and $\Gamma$ is the discrete kernel. And by a well known fact about cocompact discrete subgroups of euclidean spaces, it follows $\Gamma \simeq \mathbb{Z}^k$.