This is off the cuff, and I'm very much a dilettante when it comes to group theory, so I hope there isn't an error in what follows. Corrections welcome, of course.
[EDIT: it has been pointed out below that the group given below doesn't quite work. I'm leaving the bulk of this "answer" here, in case it suggests a correct solution or warns people off the same mistake I made.]
I think the group $G$ with presentation $\langle g, h | hg =gh^n \rangle$, where $n\geq 2$, will do the job, with $H$ being the group generated by $h$. [Conditions 2,5]
Elements of this group have a normal form with all the $g$s on the left and all the $h$s on the right. [EDIT: this is not quite right, one has to take care over negative powers of $g$.] Multiplying on the left or on the right by an element of $h$ should, once we bring to normal form, not change the index of $g$ in the normal form, and so there are infinitely many double $H$-cosets, taking care of Condition 4.
Also, given an element of the form $g^ah^b$ where $a\neq 0$, then some back-of-the-envelope scribbling indicates that repeated conjugation by $h$ ought to increase the absolute value of the index of $h$ in the resulting normal form, so that conjugation by $h$ cannot be an operation of finite order.
That would take care of Condition 3. [EDIT: this is incorrect/insufficient, see comments below.]
Finally, I think Condition 1 should follow from some further case-by-case analysis (given a non-identity element in $H$, conjugate repeatedly by $g$; and all the elements in $G-H$ are taken care of by condition 3).
(The group $G$ is an example of a Baumslag-Solitar group, and these beasts have been quite well studied over the years, I'm told. I don't know if you can do similar games with other B-S groups.)