Why does the inverse of surreal numbers exist?

[I have the 2nd edition, not the 1st, but there don't seem to be any differences important for this question.]

First, an important correction. You write "the existence of the multiplicative inverse has been established", but Conway is not (yet) claiming this. He is only claiming that the cancellation law $xy = xz \implies y = z$ (when $x \ne 0$) holds.

The cancellation law is not true in all rings, so we need something more to be able to prove it. In this case, we can use the inequality laws. Note that we have the principle of trichotomy (exactly one of $x < y$, $x = y$, and $x > y$ holds). This follows from Theorem 2 and the definitions of $=$ and $<$.

Using Theorem 8(iii), we have, if $0 < x$ and $y < z$, the conclusion $xy < xz$. Also, if $0 < x$ and $z < y$, then $xz < xy$. Applying trichotomy, this tells us that if $0 < x$ and $y \ne z$, then $xy \ne xz$. In a similar fashion, we can prove that if $x < 0$ and $y \ne z$, then $xz \ne xz$. Again applying trichotomy, if $x \ne 0$ and $y \ne z$, then $xy \ne xz$, which is equivalent to the cancellation law.

The hint from the colleague seems to suffice to prove the existence cancellation law: Theorem 9 states that $x,y>0 \Rightarrow xy >0$ which implies $xy \neq 0$ from the (surreal) definition of $>$.

Since $-xy\equiv (-x)y\equiv x(-y)$, we have e.g. $x > 0, y < 0 \Rightarrow xy \neq 0$, because would the product be equal to $0$ so would $x(-y) \equiv -xy = -0 = 0$, but $x,-y > 0$. By symmetry we have lemma: $x,y \neq 0 \Rightarrow xy \neq 0$.

Now given $xy = zy$ and assume $y\neq 0$. Set $w\equiv x-z$, then $x=z+w$ (additive inverse). By distributive law and 8 (ii) we have $$xy=(z+w)y=zy+wy$$ which can be simplified to $0=wy$. Since $y\neq 0$ we must have $w=0$ by our lemma above, hence $x=z$.