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Topological connected eccentrics, not homeomorphic to commutative Lie groups

An eccentric is a universal algebra $\ (X\ \sigma\ \lambda\ \rho)\ $ such that operations $\ \sigma\ \lambda\ \rho\,:\,X\times X\to X\ $ satisfy:

  1. $\quad \forall_{x\ y\,\in X}\quad \lambda(\sigma(x\ y)\ y)\ =\ x; $
  2. $\quad \forall_{x\ y\,\in X}\quad \rho\,(x\ \sigma(x\ y))\ =\ y; $

Let $\ \mathcal T\ $ be a topology in $\ X\ $ such that its three operations are continuous. Furthermore, let topological space $\ \mathbf X:=(X\,\ \mathcal T)\ $ be Hausdorff and connected.

Question 1:   can $\ \mathbf X\ $ be a manifold that is not homeomorphic to any Lie group? Or can $\ \mathbf X\ $ be homeomorphic to a non-commutative Lie group?

Question 2:   can $\ \mathbf X\ $ be homeomorphic to Hilbert cube or Knaster pseudo-arc?

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Examples:

Topological spaces $\,\ \Bbb R^A\times(\Bbb R/\Bbb Z)^B,\ $ for arbitrary sets $\ A\ B\ $ (possibly empty), are manifolds which admit Abelian group structure; all such Abelian topological groups are topological eccentrics. When $\ A=\emptyset\ $ we get compact manifolds called tori.

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EDIT

see my EDIT in Weirdos but algebraic.

This thread is ok as it is but in fact, I meant also to have properties:

  • $\ \forall_{a\ b\,\in\,X}\quad \sigma(\lambda(b\ a)\ a)\ =\ b;$
  • $\ \forall_{a\ b\,\in\,X}\quad \sigma(a\ \rho(a\ b))\ =\ b.$
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