The answer is: the requested colouring is not possible. Assume by contrary that such a colouring is possible. Consider the numbers are coloured in: red, green, blue. Denote: $A=\{x\in\mathbb{N}\setminus\{1\}|x\text{ is red}\}$; $B=\{y\in\mathbb{N}\setminus\{1\}|y\text{ is green}\}$; $C=\{z\in\mathbb{N}\setminus\{1\}|z\text{ is blue}\}$. From the initial conditions results: $A\cup B\cup C=\mathbb{N}\setminus\{1\};\;A\cap B=A\cap C=B\cap C=\phi$; $x\in A,\;y\in B\Longrightarrow xy\in C;\;x\in A,\;z\in C\Longrightarrow xz\in B;\;y\in B,\;z\in C\Longrightarrow yz\in A$. $\textbf{Claim 1: }$ Let be $p\in A$ a prime number. Then $p^n\in A,\;\forall n\in\mathbb{N}$. Proof: Assume exists $k\in\mathbb{N}\setminus\{1\}$ such that $p^k\in B$. $p\in A,\;p^k\in B\Longrightarrow p^{k+1}\in C$; $p\in A,\;p^{k+1}\in C\Longrightarrow p^{k+2}\in B$. By induction: $p^m\in B\cup C,\;\forall m\ge k\quad(1)$. On the other hand: $p^k\in B,\; p^{k+1}\in C\Longrightarrow p^{2k+1}\in A$, contradiction with (1). Hence, the assumption is false and results the claim. $\textbf{Claim 2: }$ Let be $p\in A$ a prime number. Then $q^n\in A,\;\forall q$ prime number and $\forall n\in\mathbb{N}$. Proof: Assume exists the prime number $q$ such that $q\in B$. $p\in A,\;q\in B\Longrightarrow pq\in C$; $p\in A,\;pq\in C\Longrightarrow p^2q\in B\quad(2)$; $p\in A\Longrightarrow \text{ (Claim 1) }p^2\in A$; $p^2\in A,\;q\in B\Longrightarrow p^2q\in C$, contradiction with (2). Hence, the assumption is false and results $q\in A,\;\forall q$ prime number. Applying the Claim 1, results: $q^n\in A,\;\forall q$ prime number and $\forall n\in\mathbb{N}$. Using the claims 1 and 2, results: all prime numbers and their powers belong to the same set $A,B$ or $C$. Assume $p^n\in A,\;\forall p$ prime number and $n\in\mathbb{N}$ and let be $N\in B$. Results: $\exists m\in\mathbb{N}\setminus\{1\};\;p_1,p_2,\dots,p_m$ pairwise distinct prime numbers$;\;\alpha_1,\alpha_2,\dots,\alpha_m\in\mathbb{N}$ such that $N=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot...p_m^{\alpha_m}$. $p_1\in A,\;p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot...p_m^{\alpha_m}\in B\Longrightarrow p_1^{\alpha_1+1}\cdot p_2^{\alpha_2}\cdot...p_m^{\alpha_m}\in C$; $p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot...p_m^{\alpha_m}\in B,\;p_1^{\alpha_1+1}\cdot p_2^{\alpha_2}\cdot...p_m^{\alpha_m}\in C\Longrightarrow p_1^{2\alpha_1+1}\cdot p_2^{2\alpha_2}\cdot...p_m^{2\alpha_m}\in A\quad(3)$. On the other hand: $p_1^{\alpha+1}\in A,\;p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot...\cdot p_m^{\alpha_m}\in B\Longrightarrow p_1^{2\alpha_1+1}\cdot p_2^{\alpha_2}\cdot...\cdot p_m^{\alpha_m}\in C$; $p_2^{\alpha_2}\in A,\;p_1^{2\alpha_1+1}\cdot p_2^{\alpha_2}\cdot...\cdot p_m^{\alpha_m}\in C\Longrightarrow p_1^{2\alpha_1+1}\cdot p_2^{2\alpha_2}\cdot p_3^{\alpha_3}\cdot...\cdot p_m^{\alpha_m}\in B$. Using successively $p_3^{\alpha_3},...,p_m^{\alpha_m}\in A$ and the partial products are elements of $ B\cup C$, we obtain finally $p_1^{2\alpha_1+1}\cdot p_2^{2\alpha_2}\cdot...\cdot p_m^{2\alpha_m}\in B\cup C$, contradiction with (3). Conclusion: It is not possible to colour all integers greater than $1{}$ in three colours such that the colour of the product of any two differently coloured numbers is different from the colour of each of the factors.

Fix a number $x$. Say $x$ is red. Then if some $x^k$ is blue, after that $x^{k+1}$ is green, $x^{k+2}$ is blue again, and so on. Likewise if some $x^k$ is green, you get $x^{k+1}$ is blue and it alternates. So if some $x^k$ is not red, eventually they alternate between blue and green. This is a contradiction, as for large $n$, we have $x^n, x^{n+1}$ are blue and green in some order but $x^{2n+1}$ is blue or green and is thus not red. So if $x$ is some color, then all powers of $x$ are also that color. In particular, $2$ and $4$ are the same color. WLOG it's red. Now, take $x$ that is not red, WLOG it is blue, then $2x$ is green, and $4x$ must be blue (because of $2x$ and $2$) and also green (because of both $x$ and $4$). We are done.