Let $G$ be a finite group, $k$ a field and $k[G]$ the group algebra. Assume I have in $k[G]\otimes k[G]$ the equality
$$\forall x\in G:\quad\sum_{g,h\in G}r_{g,h}\cdot xg\otimes xh=\sum_{g,h\in G}r_{g,h}\cdot gx\otimes hx$$
where $r(g,h)\in k$ are some coefficients, why can I conclude $r(xg,xh)=r(gx,hx)$ for all $x\in G$?
This is the usual trick of "changing coordinates". Fix $x$. Since we are summing over all elements of $G\times G$ and since multiplication by any element $y$ of $G$ gives bijections
$$G \times G \to G \times G, (g,h) \mapsto (yg,yh),$$ $$G \times G \to G \times G, (g,h) \mapsto (gy,hy),$$
we can rewrite the above equality as
$$\sum_{g,h} r(x^{-1}g,x^{-1}h) \ g \otimes h = \sum_{g,h} r(gx^{-1},hx^{-1}) \ g \otimes h,$$
where we have used both bijections (the first on the LHS and the other on the RHS) with $y=x^{-1}$.
Then you can subtract one sum from the other and you get
$$\sum_{g,h} \left[r(x^{-1}g,x^{-1}h)- r(gx^{-1},hx^{-1}) \right] g \otimes h = 0,$$
so that we have $r(x^{-1}g,x^{-1}h) = r(gx^{-1},hx^{-1})$ for all $g,h \in G$. Choosing $x^{-1}$ instead of $x$ at the beginning, we get exactly the formula you want.