This is not a solution, but ideally a clarification of the question. The semigroup $\mathbb{Z}_+$ under addition is equivalent to the semigroup $\mathbb{Z}_{\ge 0}^\infty$ under addition. We are interested in certain special subsets $A_n$ of the latter, which in multiplicative form correspond to $\{1,2,\ldots,n\}$. Each $A_n$ is a certain $n$-element lower set in the natural partial ordering on $\mathbb{Z}_{\ge 0}^\infty$. We would like to color $\mathbb{Z}_{\ge 0}^\infty$ with $n$ colors such that each translate of $A_n$ uses $n$ different colors.
By a compactness argument, we can find a sequence of translates of $\mathbb{Z}_{\ge 0}^\infty$ in $\mathbb{Z}^n$ such that the colorings converge to a coloring of all of $\mathbb{Z}^\infty$. In other words, in multiplicative form, we could equally well have asked the question in $\mathbb{Q}_+$ as in $\mathbb{Z}_+$.
Also, I think that the subset of $\mathbb{Z}^\infty$ with color 1 is a set of vectors that yields a tiling of $\mathbb{Z}^\infty$ by translates of $-A_n$. For instance, if two of these translates intersect at $x$, then you correspondingly $x+A_n$ contains two elements with color 1. (Technically this only shows that it is a packing, not a tiling. However, the cracks that are not tiled must have density 0, and you can translate them away by the same compactness argument.) Moreover, if you have such a tiling, then you can color each element of each tile in the same way, and then for that coloring each translate $x+A_n$ has all different colors. Again, in multiplicative form, you're tiling the positive rational numbers by arithmetic progressions $a,2a,3a,\ldots,na$.
So I think that the question is equivalent to tiling $\mathbb{Z}^\infty$ by translates of $A_n$. Finally, if there are $\pi(n)$ primes that are at most $n$, you might as well tile $\mathbb{Z}^{\pi(n)}$ rather than $\mathbb{Z}^\infty$. I don't know how to do this in general (unless an argument such as Victor's works), but arguably this tiling problem is a clearer view of the original question. For instance, it seems possible that the tiling can be periodic with one tile in each period, and with quotient group $\mathbb{Z}/n$. This last remark is a way to express the "linear coloring" trick in the statement of the question. As discussed in the comments, it is possible when $n+1$ or $2n+1$ is prime.