Let $R$ be a ring with unity $1_R$ and $S=\{a_1,a_2,\ldots,a_n\}$ be a subset of $R$ containing all elements of $R$ with property $P$. Define $I:=\langle\{a_1,a_2,\ldots,a_n\}\rangle$ to be the ideal generated by the set of elements with property $P$. Here $I$ is the minimum ideal containing all elements of $R$ with property $P$. The following question:
(1) If $b\in I$ and $b$ has no property $P$, is it okay for me to conclude that $b=0$?
(2) For any ideal $J$ of $R$. If $J\cap I\neq 0$, is it okay to conclude that $J\cap I=I$ by minimality of $I$ with respect to property $P$ in $R$?
(1) No it's not, why would it ? You're asking if $\langle S\rangle = S\cup \{0\}$, which is of course not true for general $S$.
(2) No of course it's not, why would it ? $J$ need not contain all the elements with property $P$.