I don't have a solution, but here are some thoughts which might be of use or interest.

You may have seen this already, but if your group is discrete then its group von Neumann algebra $VN(G)$ is "directly finite" - that is, every left invertible element is invertible. I think this property is inherited by the algebra obtained when one compresses by an idempotent in $C_c(G)$.

The earliest reference I know of is somewhere in Kaplansky's Fields and Rings; a proof of something slightly weaker, which can in fact be boosted to prove the original result, was given in

Montgomery, M. Susan. Left and right inverses in group algebras. Bull. Amer. Math. Soc. 75 1969 539--540. MR0238967 (39 #327)

(The proof needs uses the existence of a faithful tracial state on $VN(G)$, plus the fact that every idempotent in a $C^*$-algebra is similar in the algebra to a self-adjoint idempotent , -- something which was not all that obvious to me the first time I saw this result.)

I don't know what the state of play is for algebras of the form $H$, as described in your question. I think enough is known about $C^*$-algebras of some totally disconnected groups (work of Plymen et al.) that one might have similar results, but the arguments have to be different from the discrete case because one no longer has the faithful positive trace that is used by Kaplansky and Montgomery's arguments.

Of course in the more special case you have at hand, we might have enough structure to force left-invertibles to be right-invertible; but off the top of my head nothing comes to mind.

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You may have seen this already, but if your group is discrete then its group von Neumann algebra $VN(G)$ is "directly finite" - that is, every left invertible element is invertible. The earliest reference I know of is somewhere in Kaplansky's Fields and Rings; a proof of something slightly weaker, which can in fact be boosted to prove the original result, was given in

Montgomery, M. Susan. Left and right inverses in group algebras. Bull. Amer. Math. Soc. 75 1969 539--540. MR0238967 (39 #327)

(The proof needs the fact that every idempotent in a $C^*$-algebra is similar in the algebra to a self-adjoint idempotent, something which was not all that obvious to me the first time I saw this result.)

I don't know what the state of play is for algebras of the form $H$, as described in your question. I think enough is known about $C^*$-algebras of some totally disconnected groups (work of Plymen et al.) that one might have similar results, but the arguments have to be different from the discrete case because one no longer has the faithful positive trace that is used by Kaplansky and Montgomery's arguments.

Of course in the more special case you have at hand, we might have enough structure to force left-invertibles to be right-invertible; but off the top of my head nothing comes to mind.