Cancellation in a group: $gx=gy$ or $xg=yg$ forces $x=y$; equivalently left and right translation by $g$ are bijections of $G$, so $gx=h$ and $xg=h$ each have exactly one solution

Suppose $gx=gy$. Multiplying on the left by $g^{-1}$ and regrouping, $x=ex=(g^{-1}g)x=g^{-1}(gx)=g^{-1}(gy)=(g^{-1}g)y=ey=y$.

Suppose $xg=yg$. Multiplying on the right by $g^{-1}$ and regrouping, $x=xe=x(g{g}^{-1})=(xg)g^{-1}=(yg)g^{-1}=y(g{g}^{-1})=ye=y$.

For every $x\in G$, ${\lambda }_{g^{-1}}({\lambda }_g(x))=g^{-1}(gx)=(g^{-1}g)x=ex=x$, and ${\lambda }_g({\lambda }_{g^{-1}}(x))=g(g^{-1}x)=(g{g}^{-1})x=ex=x$.

For every $x\in G$, ${\rho }_{g^{-1}}({\rho }_g(x))=(xg)g^{-1}=x(g{g}^{-1})=x$, and ${\rho }_g({\rho }_{g^{-1}}(x))=(x{g}^{-1})g=x(g^{-1}g)=x$.

Existence in claim 3: $g(g^{-1}h)=(g{g}^{-1})h=eh=h$, so $x=g^{-1}h$ solves $gx=h$; and $(h{g}^{-1})g=h(g^{-1}g)=he=h$, so $x=h{g}^{-1}$ solves $xg=h$.

Claim 1 is steps 1.1 and 1.2.

By step 1.3 the map ${\lambda }_{g^{-1}}$ is a two-sided inverse of ${\lambda }_g$, so ${\lambda }_g$ is a bijection of $G$ with inverse ${\lambda }_{g^{-1}}$; by step 1.4 the same holds for ${\rho }_g$ with inverse ${\rho }_{g^{-1}}$. This is claim 2.

Uniqueness in claim 3: if $gx=h=g{x}^{\mathrm{\prime }}$ then $x=x^{\mathrm{\prime }}$ by step 1.1, and if $xg=h=x^{\mathrm{\prime }}g$ then $x=x^{\mathrm{\prime }}$ by step 1.2.

Cancellation holds on both sides, both translations are bijections, and each of the two equations has exactly one solution.