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.