Let $M=\langle S\mid R\rangle$ be a presentation for a monoid. $R=\{a_1=b_1,...\}$ is a collection of equalities, and $M$ is defined as the set of equivalence classes under the the reflexive and transitive closure of the symmetric relation of rewriting $sa_it$ as $sb_it$ and vice versa.
If for any $x\in S$ there is some $I_x$ in the free monoid $S^*$ and some relation $(xI_x=e)\in R$, then how do we prove that $M$ is a group with group presentation $\langle S\mid R\rangle$?
On
Consider some such $X, R$, and consider some monoid/group $G$ and some inclusion function $i : X \to G$. We say $i$ comports with $R$ if for all equations $T_1(x_1, \ldots, x_n) = T_2(x_1, \ldots, x_n)$ in $R$, we have $T_1(i(x_1), \ldots, i(x_n)) = T_2(i(x_1), \ldots, i(x_n))$.
Recall that the monoid/group with presentation $\langle X \mid R \rangle$ is the initial function $i \to M$ which comports with $R$. More formally, this is a monoid/group $M$, together with a canonical function $i : X \to M$ which comports with $R$, which has the following universal property:
For all monoids/groups $G$ and all $j : X \to G$ which comports with $R$, there is a unique monoid/group homomorphism $f : M \to G$ such that $f \circ i = j$.
Note that this characterises $(M, i)$ uniquely up to unique isomorphism.
Critically, note that if $M$, $G$ are groups, a group homomorphism $M \to G$ is exactly a monoid homomorphism $M \to G$. Thus, it is easy to see that if $M, i$ is the monoid with presentation $\langle X \mid R \rangle$, and if moreover $M$ is a group, then $M, i$ also satisfies the universal property which characterises being the group with the presentation $\langle X \mid R \rangle$.
So yes, $M$ can indeed be given by this group presentation.
After thinking for quite some time, I solved my question. For $w_1,w_2\in S^*$, write $w_1\sim w_2$ if it is possible to rewrite $w_1$ as $w_2$.
Proof. For any $w\in S^*$, write $w=x_1x_2...x_n$, where each $x_i\in S$, and define $I_w=I_{x_n}...I_{x_2}I_{x_1}$. The following calculation shows that $wI_w\sim e$. \begin{align*} (x_1x_2...x_n)(I_{x_n}...I_{x_2}I_{x_1})&=(x_1...x_{n-1})(x_nI_{x_n})(I_{x_{n-1}}...I_{x_1})\\ &\sim(x_1...x_{n-1})e(I_{x_{n-1}}...I_{x_1})\\ &=(x_1...x_{n-2})(x_{n-1}I_{x_{n-1}})(I_{x_{n-2}}...I_{x_1})\\ &\sim...\\ &=x_1I_{x_1}\sim e. \end{align*} Now let $f$ be the natural monoid homomorphism $S^*\to M$. For every $a\in M$, we can write $a=f(w)$ where $w\in S^*$. Then $f(w)f(I_w)=f(wI_w)=f(e)=e$, thus $a$ has a right inverse. Since $a$ was arbitrary, every element has a right inverse, so $M$ is a group. Thus, we can extend $f$ to a group homomorphism $g:F_S\to M$, such that $f$ and $g$ agree on $S^*$.
Define $R'=\{w_1w_2^{-1}\in F_S:(w_1=w_2)\in R\}$. Then by definition, $\langle S\mid R\rangle$ (as a group presentation) is equal to the quotient of $F_S$ by the normal closure $N$ of $R'$. So now our goal is to show that $\ker g=N$, after which we will be done by the First Isomorphism Theorem.
Consider any $w_1w_2^{-1}\in R'$. We know that $w_1\sim w_2$, thus $$g(w_1w_2^{-1})=g(w_1)g(w_2)^{-1}=f(w_1)f(w_2)^{-1}=f(w_2)f(w_2)^{-1}=e.$$ Hence $w_1w_2^{-1}\in\ker g$. Since $w_1w_2^{-1}$ was arbitrary, $R'\subseteq\ker g$. Since $\ker g$ is normal, $N\subseteq\ker g$ as well.
Conversely, consider any $w\in\ker g$, that is, $g(w)=e$. We want to show that $w\in N$, that is, $w$ can be obtained via inversion, multiplication by elements in $R'$ and conjugation. Write $w=y_1y_2...y_n$, where each $y_i$ is either a generator or the inverse of a generator. Next, define $z_i=y_i$ if $y_i$ is a generator, and $I_{y_i^{-1}}$ if $y_i^{-1}$ is a generator, so that $g(y_i)=f(z_i)$. Let $w'=z_1...z_n\in S^*$. Thus, $$f(w')=f(z_1...z_n)=f(z_1)...f(z_n)=g(y_1)...g(y_n)=g(y_1...y_n)=g(w)=e.$$ So $w'\sim e$. We know that a sequence of rewrites turns $e$ into $w'$, so we follow them. In each case, $sw_1t$ is rewritten as $sw_2t$, where either $w_1w_2^{-1}\in R'$ or $w_2w_1^{-1}\in R'$. In both cases, we can use $w_1w_2^{-1}$. So $$sw_2t=s((w_1w_2^{-1})^{-1}*s^{-1}(sw_2t)s)s^{-1}$$ We have used inversion, multiplication of elements in $R'$ and conjugation to get from the identity to $w'$. Therefore, $w'\in N$. Now, we use these actions to obtain $w$ from $w'$. Starting from $w'$, we successively change $z_i$ of the form $I_{y_i^{-1}}$ to the form $y_i$, like so: $$Z_1y_iZ_2=Z_2^{-1}(Z_2(Z_1I_{y_i^{-1}}Z_2)Z_2^{-1}*(y_i^{-1}I_{y_i^{-1}})^{-1})Z_2$$ Therefore, $w\in N$ and we are done.