To answer the first question: Only when $G$ is abelian. In fact, if $G$ is not abelian and has dimension $2d$, then there is no element of $H^2(G,\mathbb{R})$ whose $d$-th power is nonzero in $H^{2d}(G,\mathbb{R})$ (just look at the representation by bi-invariant forms), and there would have to be such an element if there were going to be even a symplectic structure on $G$, much less a Kähler structure. Conversely, if $G$ is abelian, then, yes, since $G=\mathbb{R}^{2d}/\mathbb{Z}^{2d}$, and, for any left-invariant complex structure, $G$ is $\mathbb{C}^d/\Lambda$, where $\Lambda\subset\mathbb{C}^d$ is a lattice (i.e., a discrete subgroup of maximal rank $2d$). Thus, there is a left-invariant Kähler structure, for example, (though there are others).

*Added on 21 Jan 2017:* As for your second question, I believe I have a partial answer, which is 'If the center of $G$ is finite (i.e., $G$ is semisimple), then $G$ does not possess a balanced metric'. Here is why: It is my understanding that, if $G$ is a compact Lie group and $J$ is a left-invariant complex structure, then $J$ is one of the complex structures constructed by Samelson and Wang. (See H.C. Wang, *Closed Manifolds with Homogeneous Complex Structure*, American Journal of Mathematics **76** (1954),1–32, which seems to indicate this though it doesn't state it explicitly. Also, see the article mentioned by François in the comments below.)

In such cases, there exists a maximal toral subgroup $T\subset G$ that is $J$-complex, and the quotient $G/T$ inherits a complex structure such that the quotient map $\pi:G\to G/T$ is holomorphic. Moreover, $G/T$ is not only Kähler, but an algebraic complex manifold to boot (in fact, it's the complete flag variety of $G$).

In particular, $G/T$ contains an algebraic (complex) hypersurface $Z\subset G/T$, whose preimage $X=\pi^{-1}(Z)\subset G$ is therefore a compact complex hypersurface in $G$. Now if $\omega$ were a positive $(1,1)$-form on $G$ (with respect to the complex structure $J$) such that $\omega^{d-1}$ were closed (where $2d$ is the dimension of $G$), then the integral of $\omega^{d-1}$ over $X$ would be positive. In particular, it would follow that the homology class of $X$ in $H_{2d-2}(G,\mathbb{R})$ were nonzero.

However, when $G$ is semi-simple (i.e., when its center is finite), we know by the Whitehead Lemma that $H_2(G,\mathbb{R}) = 0$, so Poincaré Duality implies that $H_{2d-2}(G,\mathbb{R})=0$ as well, contradicting the fact that the real homology class of $X$ must be nonzero.

I believe a more careful analysis of the cohomology ring of $G$ when the center of $G$ has positive dimension will lead to the result that $G$ can carry a balanced metric only when $G$ is abelian.