Let $G \colon = \langle x, y \ | \ x^2 = y^n = e, \ x^iy^j = x^{i^{\prime}} y^{j^{\prime}} \ $ if and only if $ i = i^{\prime}, j = j^{\prime}, \ xy = y^{-1}x \rangle$.
That is, let $G$ be the set of all formal symbols $x^iy^j$, $i = 0, 1$, $\ j = 0, 1, \ldots, n-1$, where we assume that
$\ x^iy^j = x^{i^{\prime}} y^{j^{\prime}} \ $ if and only if $ i = i^{\prime}$, $\ j = j^{\prime}$;
$x^2 = y^n = e$, where $e$ is the identity element and $n >2$; and
$xy = y^{-1}x$.
Then how to express the product $(x^iy^j)(x^ky^l)$ as $\ x^{\alpha}y^{\beta}$?
You get $yx=xy^{-1}$ from $xy=y^{-1}x$. You can prove, by induction, that $y^jx=x y^{-j}$. Since $x^2=e$, you may assume that the exponent of $x$ is always $0$ or $1$. This should be enough to express the product as you want.
The general idea (which doesn't work in general, see wikipedia) is to try "to rearrange" the symbols which are not in the right order using the relations.