You are misinterpreting the question. It's not to find an $x$ for which $10x$ is the answer to $1\bmod 9$, but rather, to find an $x$ such that the remainder when dividing $10x$ by $9$ is $1$.
All of this is much easier in modular arithmetic than working with the $\bmod$ operator. There, you problem is $10x\equiv 1 \pmod{9}$ and since $10\equiv 1 \pmod{9}$, it is directly equivalent to $x\equiv 1\pmod{9}$.
(In modular arithmetic, $a\equiv b\pmod{m}$ means that $m$ divides $a-b$; or equivalently, that the remainders when dividing $a$ by $m$ and when dividing $b$ by $m$ are the same).
Or if you want to use mod operator, because $10\bmod 9 = 1$, and $ab\bmod 9 = (a\bmod 9)(b\mod 9) \bmod 9$, then it follows that $10x\bmod 9 = x\bmod 9$. So $10x=1\bmod 9$ if and only if $x=1\bmod 9$.
However, it is not true that $x=1$ is the only answer. $x=10$ works just as well, as does $x=19$, $x=28$, $x=37$, etc.