Now I have understood the idea of Bruck's proof (hidden behind autotopies). It is simple and can be written down without involving the "heavy artillery" of autotopies.
At first, one have to prove that both (B) and (M) imply the inverse property.
Let us recall that a loop $X$ has the inverse property if for every $x\in X$ there exists an element $x^{-1}\in X$ such that $x^{-1}(xy)=y=(yx)x^{-1}$ for every $y\in X$. Writing the latter equality for $y=e$, the neutral element of the loop $X$, we see that $x^{-1}x=x^{-1}(xe)=e=(ex)x^{-1}=xx^{-1}$, which means that $x^{-1}$ is the two-sided inverse of $x$.
We shall exploit the following well-known property of loops with the inverse property.
Lemma 1. If $X$ is a loop with the inverse property, then $(xy)^{-1}=y^{-1}x^{-1}$ for every $x,y\in X$.
Proof. The inverse property implies $x^{-1}(xy)=y$, $$x^{-1}=(x^{-1}(xy))(xy)^{-1}=y(xy)^{-1}$$ and finally
$$(xy)^{-1}=y^{-1}(y(xy)^{-1})=y^{-1}x^{-1}.\quad\square$$
Lemma 2. If a loop $X$ satisfies the Bol condition (B), then it has the inverse property.
Proof. Since $X$ is a loop, for every $x\in X$ there exists a unique element $x^{-1}\in X$ such that $x^{-1}x=e$, where $e$ is the identity of the loop $X$. For every $y\in X$, the Bol condition (B) implies
$$x^{-1}(xy)=x^{-1}(x(ye))=((x^{-1}x)y)e=(ey)e=y,$$
which is the ``left'' part of the inverse property. For $y=x^{-1}$, the latter equality implies $x^{-1}(xx^{-1})=x^{-1}=x^{-1}e$ and hence $xx^{-1}=e$, by the cancellativity of the binary operation in the loop $X$. Therefore, $x^{-1}$ is the unique two-sided inverse of $x$ in the loop $X$.
To prove the ``right'' part of the inverse property, we need to check that $(zx)x^{-1}=z$ for every elements $x,z\in X$. Since $X$ is a quasigroup, there exists a unique element $y\in X$ such that $z=xy$. By the property (B),
$$z=xy=x(ye)=x(y(xx^{-1}))=((xy)x)x^{-1}=(zx)x^{-1}.$$ Therefore, the loop $X$ has the inverse property. $\square$
A loop $X$ is defined to be flexible if $x(yx)=(xy)x$ for all $x,y \in X$.
Lemma 3. If a loop $X$ satisfies the Moufang condition (M), then it is flexible.
Proof. Writing the Moufang condition (M) for $z=e$, the neutral element of the loop $X$, we obtain the identity
$$(xy)x=(xy)(ex)=((xy)e)x=(xy)x,$$
witnessing thet $X$ is flexible. $\square$
Lemma 4. If a loop $X$ satisfies the Moufang condition (M), then it has the inverse property.
Proof. Let $e$ be the identity of the loop $X$ and assume that $X$ satisfies the Moufang condition (M). Since loop $X$ is a quasigroup, for every element $x\in X$, there exists a unique element $x^{-1}\in X$ such that $x^{-1}x=e$. By the Moufang identity (M), for every $x,y\in X$, we have
$$yx^{-1}=e(yx^{-1})=(x^{-1}x)(yx^{-1})=(x^{-1}(xy))x^{-1},$$which implies $y=x^{-1}(xy)$, by the cancellativity of the binary operation in loops. The latter identity witnesses that $X$ satisfies the ``left'' part of the inverse property. For $y=x^{-1}$, we obtain $x^{-1}e=x^{-1}=x^{-1}(xx^{-1})$ and hence $e=xx^{-1}$, by the cancelativity of the binary operation in the loop $X$. Therefore, $x^{-1}$ is the two-sided inverse to $x$ in the loop $X$.
By Lemma 3, the loop $X$ is flexible. By the Moufang identity (M), for every $x,y\in X$ we have
$$x^{-1}y=(x^{-1}y) e=(x^{-1}y)(xx^{-1})=x^{-1}((yx)x^{-1}),$$
which implies $y=(yx)x^{-1}$, by the cancellativity of the binary opertaion in loops. The latter identity witnesses that the loop $X$ satisfies the ``right'' part of the inverse property and hence $X$ has the inverse property. $\square$
Now we are ready to present a short proof of the equivalence of the Bol and Moufang conditions (B) and (M).
Theorem. For every loop $X$, the conditions (B) and (M) are equivalent.
Proof. Assume that $X$ satisfies one of the conditions (B) or (M). By Lemmas 2 and 4, the loop $X$ has the inverse property. By Lemma 1, the inversion $X\to X$, $x\mapsto x^{-1}$, is an involutive anti-isomorphism of the loop $X$.
Assume that $X$ satisfies the Moufang identity (M). To prove that $X$ satisfies the identity $(B)$, take any elements $x,y,z\in X$.
Consider the elements $a=(xz)^{-1}=z^{-1}x^{-1}$ and $b=y(xz)$, and observe that $ax=(z^{-1}x^{-1})x=z^{-1}$, by the inverse property, which also implies implies $y=(y(xz))(xz)^{-1}=ba$. The inverse property and the Moufang identity $(M)$ imply $$x(y(xz))=xb=((xb)(ax))(ax)^{-1}=((x(ba))x)z=((xy)x)z,$$so, the Bol condition (B) is satisfied.
Now assume that $X$ satisfies the Bol condition (B). The condition (M) will be proved as soon as we check that $(xb)(ax)=(x(ba))x$ for any elements $x,a,b\in X$. Consider the element $y=ba$. Since $X$ is a loop, there exist a unique element $z\in X$ such that $a=z^{-1}x^{-1}=(xz)^{-1}$. The inverse property ensures that $ax=(z^{-1}x^{-1})x=z^{-1}$ and $b=(ba)a^{-1}=ya^{-1}=y(xz)$. The inverse property and the identity (B) imply
$$((xb)(ax))z=((xb)(ax))(ax)^{-1}=xb=x(y(xz))=((xy)x)z=((x(ba))x)z,$$
and hence $(xb)(ax)=(x(ba))x$, by the cancellativity of the binary operation in the loop $X$. $\square$
