The free group of a group and the kernel of a canonical morphism Let $G$ be a group and $F_G$ the free group on the set $G$. Then there exists a canonical surjective morphism ${\rm can}: F_G \to G \to 1$ constructed as follows: let $(e_x)_{x \in G}$ be a copy as a set of $G$; then $F_G$ is the free group generated by $(e_x)_{x \in G}$ -- then ${\rm can}$ is the unique morphism of groups such that ${\rm can} (e_x) := x$, for all $x\in G$. 
Is the set $\{e_x e_y e_{xy}^{-1}\}$ a system of generators of ${\rm Ker} ({\rm can})$? 
 A: The answer is finally yes.
Write $$f(x,y)=e_xe_ye_{xy}^{-1};\qquad R=\{f(x,y):x,y\in G\}\subset F_G.$$ Clearly the presentation $$\langle (e_g)_{g\in G}\mid (f(x,y))_{x,y\in G}\rangle$$
is a presentation of $G$, more precisely if $\Gamma$ is the group defined by this presentation and $\bar{e}_g$ the image of $e_g$ in $\Gamma$, we have a unique homomorphism $\phi:\Gamma\to G$ mapping $\bar{e}_g\mapsto g$ for all $g$. The first claim is that $\phi$ is an isomorphism.
Proof: [added in edit] clearly $\phi$ is surjective. Now in $\Gamma$, the inverse of $\bar{e}_g$ is equal to $\bar{e}_{g^{-1}}$; in particular every element of $\Gamma$ is a product of some of the $\bar{e}_g$ (without inverses). Since in $\Gamma$ we have $\bar{e}_g\bar{e}_h=\bar{e}_{gh}$, this shows that every element of $\Gamma$ is equal to $\bar{e}_g$ for some $g\in G$ and $\bar{e}_1=1$, and it immediately follows that $\mathrm{Ker}(\phi)$ is trivial. Hence $\phi$ is an isomorphism. $\Box$
Thus, the normal subgroup generated by $R$ is equal to the kernel $K$ of the canonical homomorphism $F_G\to G$ (mapping $e_g\mapsto g$).
The other (independent) step is to show that the subgroup $L$ of $F_G$ generated by $R$ is normal in $F_G$:
Indeed, we have the formula 
$$e_zf(x,y)e_z^{-1}=f(z,x)f(zx,y)f(z,xy)^{-1},$$
showing that $e_zLe_z^{-1}\subset L$.
Now let $U$ be the set of $w\in F_G$ such that $wLw^{-1}\subset L$. To show that $L$ is normal, we have to show that $U=F_G$. We know that $U$ is a subsemigroup of $F_G$, contains the positive generators, and contains $L$. It remains to show that it contains the inverses of generators. Indeed, we have
$$e_z^{-1}=e_z^{-1}e_{z^{-1}}^{-1}e_{z^{-1}}=f(1,1)^{-1}f(z^{-1},z)^{-1}e_{z^{-1}}\in U.\quad\Box$$
We can now combine both steps: since $L$ is normal (by the second step) and contains $R$, it contains the normal subgroup generated by $R$, which by the first step equals $K=\mathrm{Ker}(\mathrm{can})$. So $L=K$.
A: Decided to add a separate answer for something mentioned in comments: the positive answer follows (almost) immediately from the Nielsen-Schreier theorem. I think it is worth it also because what one gets is a system of free generators for $\mathrm{Ker(can)}$: what that theorem actually immediately gives is that $\mathrm{Ker(can)}$ is freely generated by $e_1$, $e_xe_{x^{-1}}$ and $e_xe_ye_{xy}^{-1}$ for $x\in G\setminus\{1\}$, $y\in G\setminus\{x^{-1}\}$. This then obviously implies what we need since $e_1=e_1e_ye_{1y}^{-1}$ and $e_xe_{x^{-1}}=(e_xe_{x^{-1}}e_{xx^{-1}}^{-1})e_1$.
The theorem in full generality states that for a group $F$ freely generated by a set $X$, any subgroup $K$ of $F$ is freely generated by the set
$$
\left\{\tau(Kw)x\tau(Kwx)^{-1}\mid w\in F,x\in X\right\}\setminus\{1\}
$$
where $\tau:K\backslash F\to F$ is any section of the quotient map $F\twoheadrightarrow K\backslash F$ which gives a Schreier transversal, i. e. has the property that the image of $\tau$ is closed under taking left prefixes.
To obtain what we want just take $X=G$, $F=F_G$, $K=\mathrm{Ker(can)}$ and let $\tau: G\to F_G$ be given by $\tau(x)=e_x$ for $x\ne1$ and $\tau(1)=1$. 
For an elementary proof of the Nielsen-Schreier see e. g. "An elementary proof that subgroups of free groups are free" by Benjamin Steinberg.
A: Yes $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $
Later
After quite some effort several people managed to convince me in the comments that the question is not as trivial as my answer seemed to imply. So I felt obliged to acknowledge it.
Still let me try to give an argument which hopefully shows why I was inclined to give that particular answer.
The subgroup $K$ generated by the words $e_xe_ye_{xy}^{-1}$ is clearly in the kernel.
To show it coincides with the kernel it suffices to show that the resulting surjective map $K\backslash F_G\twoheadrightarrow G$ (from left cosets to $G$) is injective.
To show that it suffices to reduce any word from $F_G$ to a single letter using only left multiplications by elements of $K$.
For all $x,y\in G$ the subgroup $K$ contains $e_{xy}\left(e_xe_y\right)^{-1}=\left(e_xe_ye_{xy}^{-1}\right)^{-1}$;
it also contains $e_1=e_1e_1e_{1\cdot1}^{-1}$;
hence it also contains $e_{x^{-1}}e_x=\left(e_{x^{-1}}e_xe_{x^{-1}x}^{-1}\right)e_1$.
Using this, given any word in the $e_x$es, start reading it from the left and do the following:
encountering an $\left(e_x\right)^{-1}$, multiply from the left by $e_{x^{-1}}e_x$ and cancel - as a result, the leftmost $\left(e_x\right)^{-1}$ will become replaced with $e_{x^{-1}}$;
encountering an $\left(e_1\right)^{\pm1}$, multiply from the left by $\left(e_1\right)^{\mp1}$ - as a result, the leftmost $\left(e_1\right)^{\pm1}$ drops out;
encountering $e_xe_y$, multiply from the left by $e_{xy}\left(e_xe_y\right)^{-1}$ - as a result, the leftmost $e_xe_y$ becomes replaced with $e_{xy}$.
Clearly one ends up either with a single $e_x$ or with the empty word which may (if one wishes) be replaced with $e_1$.
Now let me ask - is not all this ridiculously superfluous? :D
Still slightly later
Oops it remains to explain what to do with leftmost $e_x\left(e_y\right)^{-1}$ - in that case one has to multiply from the left by $e_{xy^{-1}}e_y\left(e_x\right)^{-1}=e_{xy^{-1}}e_y\left(e_{xy^{-1}\cdot y}\right)^{-1}$.
So I actually have to admit that I cannot make it any simpler than YCor's answer.
