Let $G$ be a topological space and $m : G \times G \longrightarrow G$ be a continuous map. Let $x_0 \in G$ be such that both the maps $m (x_0, \cdot) : G \longrightarrow G$ and $m (\cdot, x_0) : G \longrightarrow G$ are homotopic to the identity map on $G.$ Let $e_{x_0}$ denote the constant map sending every element of $G$ to $x_0.$ For two loops $f, g$ in $G$ based at $x_0$ define a loop $f \otimes g$ in $G$ by $$(f \otimes g) (t) = m (f(t), g(t))$$ based at $m (x_0, x_0).$ Can we say that $(f \ast e_{x_0}) \otimes (e_{x_0} \ast g)$ is homotopic to $f \ast g,$ where $\ast$ represents the concatenation of two paths?
If I assume that $x_0$ remains fixed during the homotopy $H : G \times I \longrightarrow G$ between $m (\cdot, x_0)$ and $\text {id}$ and during the homotopy $K : G \times I \longrightarrow G$ between $m (x_0, \cdot)$ and $\text {id}$ (in which cases $m (x_0, x_0) = x_0$) then I have an affirmative answer to this question. Just define a continuous function $L : I \times I \longrightarrow G$ by $$L (s, t) = \left\{ \begin{array}{ll} H(f(2t), s) & 0 \leq t \leq \frac {1} {2} \\ K(g(2t-1), s) & \frac {1} {2} \lt t\leq 1 \\ \end{array} \right.$$
Is it true for arbitrary homotopy $H$ and $K$ which doesn't fix the point $x_0 \in G$ during the homotopy? Any help would be greatly appreciated.
Thanks for your kind attention.