I expect you are intended to mimic the proof of Theorem 7.1, which states that the standard Coxeter presentation of $S_n$ on $n-1$ generators really does present $S_n$.
Lets call the group defined by the presentation $G_n$. The assignment $s_i \mapsto (1,2,i)$ extends to a homomorphism $G \to A_n$, which is surjective, so $|G_n| \ge |A_n|$, and we just need to prove that $|A_n| \le |G_n|$.
We do that by induction on $n$, and the base case $n=3$ is easy.
For $n>3$, define $H$ to be the subgroup $\langle s_3, s_4,\ldots, s_{n-1} \rangle$ of $G_n$. Then, by induction, $|H| \le |A_{n-1}|$, so we need to prove that $|G_n:H| \le n$.
For that, we need to guess $n$ (right) coset representatives of $H$ in $G_n$, and then prove that the guess is correct. Note that the image of $H$ under the homomorphism is the stabilizer of $n$ in $A_n$. The trick in making the guess is to choose words whose images are coset representatives of this stabilizer. That is, words for which the image of $n$ is $i$ for $1 \le i \le n$.
For example, we could choose words $$g_1,g_2,\ldots,g_n = s_n, s_n^{-1},s_n^{-1}s_3,\ldots,s_n^{-1}s_{n-1}.$$ (I compose permutations from left to right.)
Now we have to prove that, for any coset rep $g_i$ and any generator $s_j$, we have $g_is_j \in \cup_{i=1}^n Hg_i.$
I am not going to do that in detail, but here are a few examples with $n=6$.
$s_6s_4 = s_4^{-1}s_6^{-1}$; $s_6^{-1}s_3s_4 = s_6^{-1}s_4^{-1}s_3^{-1} = s_4s_6s_3^{-1} = s_4s_6s_3ss_3 = s_4s_3^{-1}s_6^{-1}s_3$; $s_6^{-1}s_3s_6 = s_6s_3^{-1} = s_3^{-1}s_6^{-1}s_3$.
The answer is no. The smallest example is the presentation of the Klein $4$-group
$$\langle x,y \mid x^2 = y^2 = 1, xy=yx\rangle.$$
In any presentation $\langle S \mid R \rangle$ of this group we have $|R| \ge |S| + 1$.
I don't know whether there is an elementary proof of that. It follows from a more general result that, for a finite presentation $\langle S \mid R \rangle$ of a finite group $G$, we have $|R| \ge |S| + |d(M(G))|$, where $d(M(G))$ is the smallest number of generators of the Schur Multiplier $M(G)$ of $G$.
$M(G)$ is the (unique up to isomorphism) largest group $M$ for which there is a group $E$ with $M \le Z(E) \cap [E,E]$ and $E/M \cong G$. For the Klein 4-group, we have $|M(G)| = 2$, where, for the covering group $E$ we can take either $D_8$ or $Q_8$.
Interestingly, $Q_8$ does have a balanced presentation (i.e. $|S|=|R|$), namely $\langle x,y \mid x^2=y^2, y^{-1}xy=x^3\rangle$.
There is a lot of literature on this topic. Search for balanced presentations, and the deficiency of a presentation.
Best Answer
It can be checked that there is a homomorphism $\phi:B\to {\rm SL}(2,3)$ with $$\phi(r) = \left(\begin{array}{cc}0&1\\2&0\end{array}\right),\ \ \phi(s) = \left(\begin{array}{cc}2&0\\2&2\end{array}\right),\ \ \phi(t) = \left(\begin{array}{cc}0&2\\1&1\end{array}\right),$$ with $\phi(r^2)=\phi(s^3)=\phi(t^3) = \phi(rst) = -I_2$, in which $\phi(s)$ has order $6$, so we just need to check that $s^6=1$ in $B$.
The relations imply that $r=st$, and the element $u=rst$ is central, with $s^3=t^3=(st)^2=u$. Let $x=st$ and $y=ts$.
Then $x^2=u$ and, since $u$ is central, $y^2=tx^2t^{-1}=u$.
Then (using $t^{-1}s^{-1}t^{-1}=su^{-1}$), we have $$[x,y] = x^{-1}y^{-1}xy=u^{-2}xyxy = u^{-2}st^2s^2t^2s = ust^{-1}s^{-1}t^{-1}s = s^3 = u,$$ so, since $[x,y]$ is central, we have $u^2=[x,y]^2=[x^2,y]=[u,y]=1,$ and so $s^6=1$ as claimed.
(Note that $\langle x,y \rangle \cong Q_8$.)