@ConnorMalin: Now I understand what you're saying. I don't think the infinite telescope construction has a natural multiplication without strict commutativity either, but this does not prevent it from being homology equivalent to the group completion.

@ConnorMalin: I'm not sure I understand. Unless I'm making a mistake, its $k$-simplices are ordered sequences $[(m,m'),(m m_1,m' m_1),\ldots,(m m_1 \cdots m_k,m' m_1 \cdots m_k)$ where $(m,m') \in M \times M$ and $m_1,\ldots,m_k \in M$. This has a perfectly good left action by $M$ which for $x \in M$ takes $(m,m')$ to $(m,x m')$.

@DanRamras: Thanks! I'll check that paper out.