If $G$ is a real Lie group, then the collection of isomorphism classes of irreducible unitary reps. of $G$ is known as the unitary dual of $G$,
and the problem you ask about, namely of explicitly describing the unitary dual
of $G$, is one of the major open problems in the theory.
If $G$ is compact, then any finite-dimensional irreducible representation is unitary (you can take any positive definite inner product on the underlying vector space of the representation, and then average it over $G$), and conversely, any irreducible unitary rep. is finite-dimensional (this is part of the Peter--Weyl theorem).
Of course, the finite dimensional irreps. of $G$ are classified by usual highest weight theory, and so the problem is solved for compact $G$.
For abelian $G$, the unitary irreps. of $G$ are just characters, and so the unitary dual of $G$ coincides with its Pontrjagin dual, for which there is a
detailed theory, which is more or less completely understood.
The major open problem is the case when $G$ is semi-simple, but non-compact.
For $GL_n(\mathbb R)$ (which is not semi-simple, I guess, but is so modulo its centre) the classification is complete (it's due to David Vogan). It is known
in some other cases as well (e.g. Vogan also treated $G_2$). I think one should be able to deduce the unitary dual for $SL_n(\mathbb R)$ from the case of $GL_n(\mathbb R)$ (and I would guess this is done in the literature).
For arbitary semi-simple groups, Harish--Chandra classified the discrete series representations, which are the unitary irreps. that can be embedded into $L^2(G)$, and (building on Harish--Chandra's results) Knapp and Zuckerman classified the so-called tempered irreps. for any $G$.
There are other results known; some of the relevant names are --- in addition to Vogan --- Adams, Arthur, Barbasch, Schmid and Vilonen. However, as far as I know, for the moment the problem remains open for general semisimple groups.
If $G=H\times K$, we have $\mathrm{ind}^{H'}_H(\phi)\cdot\mathrm{ind}^{K'}_K(\psi)=\mathrm{ind}^{H'\times K'}_{H\times K}(\phi\cdot\psi)$ so property (A) holds for a product of groups if it holds for each factor. Therefore it holds for all abelian groups.
The second part of that statement is wrong. I have wrongly assumed that the product of $\mathbb{Q}$-irreducible characters is $\mathbb{Q}$-irreducible. This can fail because the Schur index of a product only divides the product of Schur indexes. The example I talked about was in the old french edition; in the 1977 english edition, it is given with more details as exercise 13.4. Let me elaborate.
Let $\phi$ denote the faithful irreducible complex character of the quaternion group $H$ and $\psi=2\phi$ the faithful $\mathbb{Q}$-irreducible character of $H$ (see Linear representation theory of quaternion group). Let $\chi$ denote the character of the faithful irreducible rational representation of $C_3$ given by $\mathbb{Q}(e^{2i\pi/3})$. Then Serre explains how to construct a faithful $\mathbb{Q}$-irrreducible representation of $H\times C_3$ with character $\phi\cdot\chi$; in particular the character $\psi\cdot\chi=2(\phi\cdot\chi)$ is not $\mathbb{Q}$-irreducible.
Moreover, one computes $<\phi\cdot\chi,\phi\cdot\chi>=2$ and $\phi\cdot\chi$ is of degree 4, hence the multiplicity of $\phi\cdot\chi$ in the regular representation is
$$
\frac{<1_{1}^{H\times C_3},\phi\cdot\chi>}{<\phi\cdot\chi,\phi\cdot\chi>}=\frac{(\phi\cdot\chi)(1)}{2}=2
$$
which implies by exercise 13.3 that $\phi\cdot\chi$ is not a $\mathbb{Z}$-linear combination of permutation characters.
For more information about which virtual $\mathbb{Q}$-characters are linear combinations of permutation characters, one can read Bartel, Dokchitser, Rational representations and permutation representations of finite groups
Best Answer
The group $GL_2(\mathbb C)$ is the product of its centre $\mathbb C^{\times}$ and its subgroup $SL_2(\mathbb C)$. This product is not direct; the two subgroups intersect in $\{\pm 1\}$ (the centre of $SL_2(\mathbb C)$).
Suppose $\pi$ is an irred. unitary rep. of $SL_2(\mathbb C)$, and let $\varepsilon$ be the character through which $\{\pm 1\}$ acts on $\pi$. Choose an extension of $\epsilon$ to a unitary character of $\mathbb C^{\times}$. Then there is a unique extension of $\pi$ to a unitary rep. of $GL_2(\mathbb C)$, on which $\mathbb C^{\times}$ acts via the character $\chi$.
Conversely, any irred. unitary rep. of $GL_2(\mathbb C)$ arises from one of $SL_2(\mathbb C)$ by such a construction.
[The case of $GL_2(\mathbb R)$ is a little more subtle, and this is what I was thinking of when I wrote my comment above. The reason is that in this case $GL_2(\mathbb R)$ is not the product of its centre and $SL_2(\mathbb R)$, because the squaring map $\mathbb R^{\times} \to \mathbb R^{\times}$ --- which is what arises from restricting the determinant map to the centre of $GL_2$ --- is not surjective, unlike in the case of $GL_2(\mathbb C)$.]