A question about dihedral group

YES, it is. Let's call $G_0$ your presentation of $D_{2n}$. The universal property of $G_0$ tells you that there is a morphism $G_0\to G$ which sends $r$ to $a$ dand $s$ to $b$.

Claim. $G$ is generated by $a$ and $b$.

For, it is enough to show that the various elements $a^k, ba^\ell$ are pairwise distinct (where $0\leq k,\ell\leq n-1$).

The only non trivial thing to prove is that $a^k= ba^\ell $ for some $k,\ell$ cannot happen. Assume the contrary, so that $b=a^m, m=k-\ell$, so $-(n-1)\leq m\leq n-1$. Taking inverse and using the fact that $b$ has order $2$, one may assume that $0\leq m\leq n-1$.

The order of $b$ is $2$, so if $n$ is odd it cannot happen. Write $n=2r$. Then $b=a^r$, the unique element of order of the cyclic group generated by $a$. Now $ba=a^{r+1}$, which has order $2r/gcd(2r,r+1)$. Now $gcd (2r,r+1)$ is $1$ or $2$ since $2(r+1)-2r=2$. Consequently, $o(ba)=2r$ or $r$.

Therefore, if $r\geq 3$ you have a contradiction. If $r=2$, o$(a^3)=4$ and you have a contradiction again.

To sum up, $G$ is generated by $a$ and $b$. Hence the morphism above is surjective, hence bijective.

The answer is yes

Proof:

Note that $[G:\langle a\rangle]=2$ so $\langle a \rangle$ is normal in $G$.

We also have $baba=1$ so $b^{-1}ab=a^{-1}$.

This means $G\cong C_n\rtimes C_2=D_{2n}$

Since in $G$, the relation $a^n=b^2=1, (ab)^2=1$ hold (i.e., the ones defining $D_{2n}$), it follows that $H:=\langle a,b\rangle$ is a subgroup of $G$ that is isomorphic to a quotient of $D_{2n}$ (via $r\mapsto a$, $s\mapsto b$). From $a\in H$, clearly $|H|\ge n$, so either $H=G$ or $H=\langle a\rangle$. In the first case, $H$ has the same (finite) order as $D_{2n}$, hence the quotient $H$ (and so $G$) must be isomorphic to $D_{2n}$ itself. In the second case, $H$ is cyclic. A cyclic group has at most one element of order $2$, hence $b=ab$, so $a=1$, contradicting $b\in\langle a\rangle$.