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This is exercise 6.17 in Steps in Commutative Algebra:

$I$ is ideal of the commutative ring $R$. Prove $I=(0:1+I)$.

I proved it wrong like this, let $r\in (0:1+I)$ then $r=r(1+0)=0$. So if the conclusion is true, then $I=0$. Did I misunderstand something?

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    $1+I$ stands for the residue class of $1$ modulo $I$. (See page 21.) – 2017-01-19

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It seems that you want to prove that $I=(\overline{0}:_R\overline{1})$, where the residue classes are taken mod $I$.

If $r\in I$, then $r\cdot \overline{1}=\overline{r}=\overline{0}$ because by hypothesis $r\in I$, then $r\in (\overline{0}:\overline{1})$. This means that $I\subseteq (\overline{0}:\overline{1})$.

If $x\in (\overline{0}:\overline{1})$, then $x\cdot \overline{1}=\overline{x}=\overline{0}$, therefore $x\in I$. This means that $(\overline{0}:\overline{1})\subseteq I$.

Hence $I=(\overline{0}:\overline{1})$.