On textbooks, the dihedral group $D_{2n}$ is written as $\langle r,s \mid r^n=s^2=1,sr=r^{-1}s\rangle$.
However, sometimes a very bad choice of $r,s$ can make the generated group be strictly smaller than $D_{2n}$, such as $r=s$ and $2|n$ (I conjecture that no matter what $r,s$ we choose, such sets should be isomorphic to a subgroup of $D_{2n}$ anyway).
Then is it itself insufficient to construct $D_{2n}$?
If then, what do we need more?
On
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$The meaning of the expression $$ G = \Span { r,s \mid r^n=s^2=1,sr=r^{-1}s } $$ is the following. Consider the free group $F$ on the two (distinct) elements $\rho, \sigma$. Let $N$ be the smallest normal subgroup of $F$ containing by $$ \rho^{n}, \sigma^{2}, \sigma \rho \sigma \rho. $$ Then $G$ is defined as $$ G = F / N, $$ with $s = \sigma N$, $r = \rho N$.
You are misunderstanding how the group is given to you: The elements $r$ and $s$ and specified elements. There are not yours to choose as you wish!
For example, if as you suggest we take the pair $(s, s)$ then these two elements generate a subgroup of order two in $D_{2n}$. But $r$ is not equal to $s$ here, because they are not equal in the group.
That said, certain choices of pairs of elements do generate the whole of $D_{2n}$. For example, if $k$ is coprime to $n$ then the pair $(r^k, s)$ generates the whole of $D_{2n}$.
On the other hand, we can "force" $r$ and $s$ to be equal. This corresponds to taking a homomorphic image. In particular, you get the group with presentation $\langle r, s\mid r^n, s^2, rs=r^{-1}s, r=s\rangle$, and if $n$ is odd then this is the trivial group. This, I believe, is what you are observing. [Here the normal subgroup $N$ such that $D_{2n}$ is the above group is the smallest normal subgroup containing the element $sr^{-1}$; this is often written as $N=\langle\langle sr^{-1}\rangle\rangle$.]