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While it's true that in $M\otimes_RN$ you can do $$ xr\otimes y=x\otimes ry $$ you can't "exchange ideals" across the tensor product. A simple example should make this clear: set $R=\mathbb{Z}$, $I=2\mathbb{Z}$ and $N=\mathbb{Z}/2\mathbb{Z}$. Since, as $\mathbb{Z}$-modules we have $\mathbb{Z}\cong 2\mathbb{Z}$, we have $$ \mathbb{Z}\otimes N\cong (2\mathbb{Z})\otimes N $$ which is isomorphic to $N$. On the other hand, $IN=0$, because $I=2\mathbb{Z}$ is precisely the annihilator of $N$. So your reasoning is faulty to begin with.

When you have doubts about tensoring, drawing a commutative diagram can help.

Consider the exact sequence $0\to I\to R\to R/I\to 0$ and tensor it with $N$: you get the diagram with exact rows $$\require{AMScd} \begin{CD} 0@>>>\mathrm{Tor}^R_1(R/I,N)@>>> I\otimes_R N@>>> R\otimes_R N@>>> (R/I)\otimes_R N@>>> 0 \\ @. @. @VVV @VVV @VVV \\ {} @. 0 @>>> IN @>>> N @>>> N/IN @>>> 0 \end{CD} $$ where $R\otimes_R N\to N$ and $(R/I)\otimes_R N\to N/IN$ are isomorphisms and $I\otimes_R N\to IN$ is surjective. This means that the kernel of $I\otimes_RN\to IN$ is isomorphic to $\mathrm{Tor}^R_1(R/I,N)$ which is not necessarily zero.

It doesn't matter whether you know about Tor; just remember that tensoring doesn't (necessarily) preserve monomorphisms, so you can simply put in the kernel $K$ of the induced morphism $I\otimes N\to R\otimes N$. You now see clearly that proving $I\otimes N\to IN$ being an isomorphism is equivalent to proving that $K=0$.

In the example, we have $IN=0$, so clearly $I\otimes N$ is not isomorphic to $IN$. Actually, the induced morphism $I\otimes N\to R\otimes N$ is the zero map.

Saying that $\mathrm{Tor}^R_1(R/I,N)=0$ for all ideals $I$ is just saying that every $R$-module is flat.