Shameless upfloat of 1-year old question - the motivation is that in general the corresponding Banach version is false, so I am trying to see where the proof breaks down, and what (if anything) can be salvaged.

Rather embarrassingly, the question is on how to prove something I used to know how to do... but I thought that the collective wisdom of MO might be able to quickly sketch how the argument should go, or direct me to a reference.

Throughout all algebras are over a field $k$ of characteristic zero (in fact ${\mathbb C})$ and have identity elements. $\newcommand{\rhc}{\overline{HC}}$

Given such $A$ and $B$, let $A\ast B$ be their free product (coproduct in the category of unital algebras). I seem to remember convincing myself, or reading, that $$ \rhc_n (A\ast B) \cong \rhc_n(A) \oplus \rhc_n(B) $$ where $\rhc$ denotes reduced cyclic homology (so $\rhc_\bullet(k)=0$ rather than having non-zero contributions in even degrees).

Unfortunately I can't remember how the proof goes. My vague recollection is that it goes via the identification of $\rhc$ as the derived functor associated to the comonad ${\rm Vect}_k \leftrightarrow {\rm Alg}_k$ (which follows from the fact that for any vector space $V$, the reduced cyclic homology of the tensor algebra $T(V)$ vanishes in all positive degrees). But how to proceed? It should be some kind of acyclic complex argument, related to the fact that $T(V_1\oplus V_2) \cong T(V_1)\ast T(V_2)$, but I can no longer recall how these kinds of argument work in detail.