Prove that $(1-\sqrt{-5})\otimes(1+\sqrt{-5})\not=2\otimes 3$ in a specific tensor product.
We have:
$$2 [(1 - \omega) \otimes (1 + \omega)] = 2(1 - \omega) \otimes (1 + \omega) = 2 \otimes (1 - \omega) (1 + \omega) = 2 \otimes 6 = 2 [2 \otimes 3].$$
Similarly,
$$3 [(1 - \omega) \otimes (1 + \omega)] = (1 - \omega) \otimes 3 (1 + \omega) = (1 - \omega) (1 + \omega) \otimes 3 = 6 \otimes 3 = 3 [2 \otimes 3].$$
Therefore, subtracting the two gives:
$$(1 - \omega) \otimes (1 + \omega) = 2 \otimes 3.$$
A possible brute force method to solve the problem would be:
First, $I$ is generated as an Abelian group by $1-\omega, \omega(1-\omega) = 5 + \omega, 2, 2\omega$. Now, you could find the relations between these elements by finding the kernel of the matrix $A = \begin{bmatrix} 1 & 5 & 2 & 0 \\ -1 & 1 & 0 & 2 \end{bmatrix}$ (as a subgroup of $\mathbb{Z}^4$). One way to do this would be to use a Smith normal form calculation to write $A = P \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix} Q$ for invertible matrices $P, Q$, and then the kernel of $A$ would be generated by $Q^{-1} e_3, Q^{-1} e_4$.
If I'm not mistaken (though I very well could be), this should give presentations $I \simeq \langle a, b \mid 2a = (1 - \omega) b, (1 + \omega) a = 3 b \rangle$ and $J \simeq \langle c, d \mid 3c = (1 + \omega) d, (1 - \omega) c = 2d \rangle$ for $I$ and $J$ as $R$-modules. Therefore, $I \otimes_R J$ would have a presentation in terms of 4 generators $a \otimes c, a\otimes d, b\otimes c, b\otimes d$, and eight relations $2a\otimes c = (1-\omega)b \otimes c, \ldots, (1-\omega)b\otimes c = 2b\otimes d$. You now want to know whether $a\otimes c - b\otimes d$ is zero in this tensor product, which is equivalent to asking whether $a\otimes c - b\otimes d$ is in the submodule of $R^4$ generated by these relations. That question should be straightforward if tedious to answer by a Smith normal form calculation on an $8 \times 16$ matrix (using that the tensor product is generated as an Abelian group by $a\otimes c, \omega a\otimes c, \ldots$ and similarly the relations are given by the original relations along with the relations times $\omega$).
In this case the natural map $I\otimes J \to IJ$, \begin{eqnarray} I\otimes J \ni i\otimes j \mapsto ij \in IJ \subset R \end{eqnarray} is an isomorphism. The reason is that $R$ is a Dedekind domain, in which every non-zero ideal is invertible, so projective, and so flat. Now it is enough to check your equality inside $R$. Your two elements both map to $6$, so they are equal.
Obs: I wonder if you can find a counterexample for some other $R$, like $R=\mathbb{Z}[\sqrt{-3}]$.
$\bf{Added:}$ Looking at the solution of @Daniel Schepler, we see how his trick would work in general.
Say we have $R$ a ring, and $a$, $b$, $c$, $d$ elements of $R$ such that $a\cdot b = c\cdot d$. Then inside $(a,c)\otimes (b,d)$ we have \begin{eqnarray} a\cdot a\otimes b = a a \otimes b = a \otimes ab = a \otimes cd = a c\otimes d = a \cdot c \otimes d \end{eqnarray} and similarly $x \cdot a\otimes b = x \cdot c \otimes d$, where $x \in \{ a,b,c,d\}$. Therefore, if the ideal $(a,b,c,d)=(1)$, the unit ideal, we can conclude $a\otimes b = c\otimes d$.