There are good answers to this question both in formal answers and in the comments, but I'll make a couple of general remarks. If a finite group $G$ has a (necessarily normal) Abelian subgroup $A$ of index $2,$ then any complex irreducible character $\chi$ of $G$ has degree at most $2,$ for by Clifford's Theorem, ${\rm Res}^{G}_{A}(\chi)$ is a sum of (necessarily degree $1$) irreducible characters of $A$. Note that $\langle {\rm Res}^{G}_{A}(\chi),{\rm Res}^{G}_{A}(\chi) \rangle_{A}\leq 2$ since $\langle \chi, \chi \rangle_{G} = 1.$
Hence ${\rm Res}^{G}_{A}(\chi)$ is either irreducible, or is a sum of two distinct irreducible characters of $A$ (in the same $G$-orbit).
Hence determining the irreducible characters of such group boils down down to determining which irreducible characters of $A$ extend to $G$ ( ie are $G$-stable). The $G$-stable irreducible characters of $A$ each extend in two ways to an irreducible character of $G$ ( given any one extension, it does not vanish identically outside $A$, and if we multiply it by the linear character of $G$ with kernel $A$, we get a different extension).
Any irreducible character of $A$ which is not $G$-stable induces irreducibly to $G$, in the same way as the other irreducible character in its $G$-orbit.
In your group, it is easy to determine which irreducible characters of $\langle a \rangle$ are $b$-stable (hence $G$-stable), so it is routine to determine the character table.