Dividing versus multiplying by inverse in modular arithmetic

Your original method works only because $10$ has an inverse modulo $11$. You wrote $20 \equiv 10x \pmod{11}$ as $$ 10(2) \equiv 10x \pmod{11}. $$ Now, using the fact that $10^{-1}$ exists, we can multiply by it to conclude $2 \equiv x \pmod{11}$. If $10$ wasn't invertible, we wouldn't be able to conclude this. For instance, $2(2) \equiv 2(3) \pmod{2}$, but $2\not\equiv 3 \pmod{2}$.

Invertible elements are always cancellable: $\,a^{-1}$ times $\,ab\equiv ac\,\Rightarrow\,b\equiv c,\,$ i.e. $\,x\mapsto ax\,$ is $1$-$1$. Furthermore: $\ a\,$ is invertible mod $m\iff a\,$ is coprime to $m\iff x\mapsto ax\,$ is onto, see

Theorem $\ $ The following are equivalent for integers $\rm\:a, m.$

$(1)\rm\ \ \ gcd(a,m) = 1$
$(2)\rm\ \ \ a\:$ is invertible $\rm\ \ \ \,(mod\ m)$
$(3)\rm\ \ \ x\to ax\:$ is $\:1$-$1\:$ $\rm\,(mod\ m),\,$ i.e. $\rm\,a\,$ is cancellable $\!\bmod m,\,$ i.e. $\rm\,ax\equiv ay\Rightarrow x\equiv y$ $(4)\rm\ \ \ x\to ax\:$ is onto $\rm\,(mod\ m)$

Proof $\ (1\Rightarrow 2)\ $ By Bezout $\rm\, gcd(a,m)\! =\! 1\Rightarrow ja\!+\!km =\! 1\,$ for $\rm\,j,k\in\Bbb Z\,$ $\rm\Rightarrow ja\equiv 1\!\pmod{\! m}$
$(2\Rightarrow 3)\ \ \ \rm ax \equiv ay\,\Rightarrow\,a(x\!-\!y)\equiv 0\,\Rightarrow\,x\!-\!y\equiv 0\,$ by multiplying by $\rm\,a^{-1}$
$(3\Rightarrow 4)\ \ $ Every $1$-$1$ function on a finite set is onto (pigeonhole).
$(4\Rightarrow 1)\ \ \ \rm x\to ax\,$ is onto, so $\rm\,aj\equiv 1,\,$ some $\rm\,j,\,$ i.e. $\,aj\!+\!km = 1,\,$ some $\rm\,k,\,$ so $\rm\gcd(a,m)=1$

See here for a conceptual proof of said Bezout identity for the gcd.

Beware if $\,a\,$ isn't invertible it fails; then $\ 1< d\mid a,m\,$ ao $\,dx\equiv 0\,$ has two roots $\,x\equiv 0,m/d\ $ so we can't cancel $\,d\,$ to deduce $\,x\equiv 0$