So, this was my initial proof:
Assume $R$ is a ring, and $a,b\in R$
Let $x_1$ and $x_2$ be solutions of $ax=b$
Hence, $ax_1=b=ax_2 \Rightarrow ax_1-ax_2=0_R \Rightarrow a(x_1-x_2)=0_R$
Thus, we have $x_1-x_2=0_R \Rightarrow x_1=x_2$, and only one solution exists.
Only now did I realize that I can only assume $x_1-x_2=0_R$ from $a(x_1-x_2)=0_R$ if $R$ was an integral domain. I didn't know why they provided that $R$ had an identity or why $a$ is a unit.
Best Answer
Your solution is close to being great. First off, they told you that $R$ is ring with unity because only those can have units. Second, your proof doesn't require knowledge that it is an integral domain. Towards the end, you have
$$a(x_1-x_2)=0$$
Then, because $a$ is a unit, $a^{-1}$ exists, and
$$a^{-1}a(x_1-x_2)=a^{-1}0=0$$
$$x_1-x_2=0\implies x_1=x_2$$
so the solution is unique, as you stated.