A projective ideal in an integral domain is finitely generated

Your approach is wrong since $R_P/I_P\otimes I_P$ is typically not $0$. For instance, if $I$ is principal, then it is isomorphic to $R$ as a module, so $R_P/I_P\otimes I_P\cong R_P/I_P\otimes R_P$.

Here is a hint for what you can do instead. Take a set $S$ of generators of $I$, which gives a surjective homomorphism $f:R^{\oplus S}\to I$. Since $I$ is projective, this maps splits via some map $g:I\to R^{\oplus S}$. Now use the fact that $R$ is a domain and $I$ is an ideal to show that the image of $g$ is nonzero on only finitely many coordinates, and so there is a finite subset $S_0\subseteq S$ which still generates $I$.

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For any $a,b\in I$, note that $$bg(a)=g(ab)=ag(b).$$ If $a$ and $b$ are nonzero, this implies that $g(a)$ and $g(b)$ are nonzero on exactly the same (finite) set of coordinates, since multiplying by $a$ or $b$ cannot change whether a coordinate is $0$ (here we use the fact that $R$ is a domain). So there is a finite set $S_0\subseteq S$ which is the set of coordinates on which $g(a)$ is nonzero for all nonzero $a\in I$. We then see that $f$ is still surjective when restricted to $R^{\oplus S_0}$, since $f\circ g=1_I$ and the image of $g$ is contained in $R^{\oplus S_0}$. Thus $S_0$ generates $I$.