Clearly, every element in $A$ must be of the same color of one of the 10 primes. Take $P=p_1p_2\cdots p_{10}$. $P$ is in $A$, and $P$ is colored in a color $C$, which, without loss of generality, is the same color as $p_1$. We will prove that every multiple of $p_1$ is of color $C$. Suppose that there exists $n=ap_1\in A$ colored in a color $C'\neq C$. If we take $P^2=P\cdot P$, then $P^2$ is of color $C$, and $P^3=P^2\cdot P$ too, as well as any power of $P$. There exists an integer $t$ such that $n\mid P^t$ Now, $p_1$, a number of color $C$, divided $n$, a number of color $C'$ which divided $P^t$,a number of color $C$. This is impossible because of the third condition.

By abuse of notation, we identify colors as subsets of $A$. Let's say color $\mathcal A$ is recessive to color $\mathcal B$ if there exists $a \in \mathcal A$ and $b \in \mathcal B$ such that $a \mid b$. We write this as $\mathcal A \lessdot \mathcal B$. Claim: [Anti-symmetry] We can't have both $\mathcal A \lessdot \mathcal B$ and $\mathcal B \lessdot \mathcal A$. Proof. This is the third condition. $\blacksquare$ Claim: If $\mathcal A \lessdot \mathcal B$, then for any $a \in \mathcal A$ and $b \in \mathcal B$, we have $ab \in \mathcal B$. Proof. By second condition, we know $ab$ is colored either $\mathcal A$ or $\mathcal B$, but the former case would cause $\mathcal B \lessdot \mathcal A$ since $b \mid ab$. $\blacksquare$ Claim: [Transitivity] If $\mathcal A \lessdot \mathcal B$, and $\mathcal B \lessdot \mathcal C$, then $\mathcal A \lessdot \mathcal C$. Proof. Suppose $a \mid b$ and $b' \mid c$ witnesses the first two relations. Now, $bb' \in \mathcal B$, and $bc \in \mathcal C$. So $a \mid bb' \mid bc$ and we get $\mathcal A \lessdot \mathcal C$. $\blacksquare$ It follows $\lessdot$ is a total order. So it has a maximal element, that is, a color $\mathcal M$ which is not recessive to any other. Let $p \in \mathcal M$ be the prime of $P$ in $\mathcal M$, promised by the first condition. It now follows all multiples of $p$ are in $\mathcal M$, via the second claim.

What a cool problem, I'm glad I was able to solve it during the competition. This solution wasn't proposed by any other participant or tutor but was the one I found and I think it's very cool and interesting. First, since every $a\in A \implies a$ can be repeatedly factored as $(p_i)(a/p_i)$ and since every $mn$ has the color of $m$ or $n\implies$ there are exactly 10 colors since there are $10$ $p_i$ $'s$ and each one of them has a different color. For the sake of contradiction let's assume that $\forall p_i \exists x\in \mathbb{Z}^+, x\geq 2$ such that $p_i$ and $xp_i$ have a different color $\implies \exists p_j$ such that $xp_i$ and $p_j$ have the same color $\implies$ we can have $p_i,yp_j$ and $xp_i,p_j$ and due to rule $c)$ $\implies p_i$ and $yp_j$ have a different color $\forall y\in A$ The possible values of $yp_j$ include $p_1p_2...p_{10} \forall p_j \implies p_i$ and $p_1p_2...p_{10}$ can't have the same color $\forall p_i\implies p_1p_2...p_{10}$ can't have a color. $\therefore$ Since $\forall a\in A$ has a color $\implies$ we have a contradiction $\implies \exists p_i$ such that all the multiples of $p_i$ have the same color. Q.E.D. I hope you like this cool and unique solution

Nice one. Say that $p_i$ dominates $p_j$ iff $p_ip_j$ is coloured $p_i$ (i.e it has the same color as $p_i$). Say that $p_i$ dominates $p_j$, which dominates $p_k$. We claim that $p_k$ is dominated by $p_i$. FTSOC the opposite holds true. Note, then we have that $p_ip_jp_k$ is either coloured $p_j$ or $p_k$. In the former case, consider $p_i,p_ip_j,p_ip_jp_k,p_k$ which contradicts the third case. In the latter, do the same, just replacing $i$ with $j$ and $j$ with $k$. Now suppose that $p_x$ dominates the maximum number of primes. If it does not dominate everyone, someone dominates it and also everyone it dominates, which is a contradiction to maximality. Now let $N$ be any combination of primes containing the prime dominating everyone else (say $p_i$). If it is not coloured $p_i$, say it is coloured $p_k$ for some $k \neq i$. It follows that $p_i,N,p_k,p_ip_k$ is a contradiction to the third condition, as desired.

First consider $P=p_1p_2\cdots p_{10}$, then suppose $P$ it's colored with $p_i$ color, it´s clear that every $p_r$ divides $P$, then take m,n,j,k to be $P,p_i,p_x,S$, with $i$ ≠ $x$ and $S$ a multiple of $p_i$ we know that if $S$ it´s colored with the color of $p_y$ and $y$ ≠$i$ we can take $p_x$ = $p_y$ and $p_x\mid P$, $p_i\mid S$ which contradicts the third condition. Finally $S$ only can be colored with $p_i $'s color and every multiple of $p_i$ it's colored with $p_i$'s color

Let $k=p_1p_2 \cdots p_{10}$ have the same color as $q \in P$. We claim $q$ is the desired prime. Indeed, assume that if $q$ divides $\ell$ then $\ell$ has the color of $r \in P$ not equal to $q$. $q$ divides $\ell$ and $r$ divides $k$ and $q$ and $k$ have the same color. Thus, $r$ and $\ell$ cannot have the same color. Contradiction $\blacksquare$

This problem has no rhyme or reason to it; it just solves itself. Obviously, $P$ must be colored with exactly $10$ colors. Let $p_i$ be colored $i$ for each $i$, and assume without loss of generality that $p_1p_2\cdots p_{10}$ is colored with $1$. Then, $(p_1p_2\cdots p_{10})^k$ is colored $1$ for every $k$ inductively by writing it as $(p_1p_2 \cdots p_{10}) \cdot (p_1p_2 \cdots p_{10})^{k-1}$, and For any $p_1 \mid r$, setting $m=n=r$, $k = (p_1p_2\cdots p_{10})^k$, and $j = p_1$ forces $r$ to be colored $1$.