There is no such set $S$. Suppose for a contradiction that there was. By rescaling the coordinates, we can assume all coefficients of points in $S$ are positive integers. Now construct a set $S'$ as follows: for every point $(a,b,c,d)\in S$, put points $(a,b,0,0), (a,0,c,0), (a,0,0,d), (0,b,c,0), (0,b,0,d), (0,0,c,d)$ into $S'$. $S'$ will then be a solution to the problem if $S$ was, so replace $S$ by $S'$.
By adding finitely many additional points to $S$, we may further assume that if we decrease a coefficient of a point in $S$ by $1$, then the resulting point is still in $S$ as long as the coefficient was greater than $1$ to start with. $S$ now has the following properties:
Property 0: Every point in $S$ has exactly two positive coefficients, and the maximal $M$ such that $(M,1,0,0)\in S$ is the same as the maximal $M'$ such that $(M',0,1,0)\in S$.
Property 1: If $(a,b,0,0)\in S$ and $(0,0,c,d)\in S$, then at least one of $(a+1,0,c+1,0), (a+1,0,0,d+1), (0,b+1,c+1,0), (0,b+1,0,d+1)$ is in $S$.
Proof: Let $A$ be maximal such that $(A,b,0,0)\in S$ and let $D$ be maximal such that $(0,0,c,D)\in S$. Some element of $S$ dominates $(A,b,c,D)$, and by the choice of $A$ (respectively $D$) it can't have its last two (respectively first two) coefficients both equal to $0$.
Property 2: If $(x,y,0,0)\in S$, then at least one of $(x+1,0,0,2), (0,y+1,0,2)$ is in $S$.
Proof: Let $X$ be maximal such that $(X,y,0,0)\in S$, and let $M$ be maximal such that $(0,0,M,1)\in S$. Then some element $(a,b,c,d)$ of $S$ dominates $(X,y,M,1)$. By Property 0 we have $c\le M$, so $c = 0$. Since we can't have $(X+1,y+1,0,0)\in S$, $d$ must not be $0$, so $d \ge 2$ and either $a \ge X+1 \ge x+1$ or $b \ge y+1$.
Now consider the following property, depending on a parameter $k$:
Property $k$: If $(x,y,0,0)\in S$, then at least one of $(x+1,0,0,k), (0,y+1,0,k)$ is in $S$.
If $S$ has Property $k$ for every integer $k$, then clearly $S$ must be infinite. We'll now prove that in fact Property $k$ implies Property $k+1$ for $k \ge 2$, and this will give our desired contradiction.
Property $k$ implies Property $k+1$: Choose $A,B,C$ maximal such that $(A,0,0,k), (0,B,0,k), (0,0,C,k) \in S$. To see that such $A,B,C$ exist at all, we apply Property $k$ to the point $(1,1,0,0)$ to see that either $(2,0,0,k)$ or $(0,2,0,k)$ is in $S$, and then we apply Property $0$ to see that all of $(1,0,0,k), (0,1,0,k), (0,0,1,k)$ are in $S$.
Now we construct a sequence of points $s_i \in S$, each with fourth coordinate equal to zero, as follows. Start with $s_0 = (x,y,0,0)$. For each $i$, split into several cases:
- If $s_i = (0,u,v,0)$, apply Property 1 to $(0,u,v,0)$ and $(A,0,0,k)$, and call the resulting dominating point $s_{i+1}$. If the fourth coordinate of $s_{i+1}$ is nonzero, stop here.
- If $s_i = (u,0,v,0)$, apply Property 1 to $(u,0,v,0)$ and $(0,B,0,k)$. Proceed similarly to the above.
- If $s_i = (u,v,0,0)$, apply Property 1 to $(u,v,0,0)$ and $(0,0,C,k)$. Proceed similarly to the above.
The claim is that this process must stop, and the final $s_{i+1}$ produced will be the point we are looking for. To see this, first note that we can never end up in the same case twice in a row during this process. Furthermore, we can never pass through all three cases in the course of this process: for instance, if we were to start in case 3 at step $i-2$, then go through case 2 at step $i-1$, and then end up in case 1 at step $i$, then if we write $s_i = (0,u,v,0)$ we find that $u = B+1$ and $v = C+2$. Applying Property $k$ to $s_i$, we see that one of $(0,B+2,0,k), (0,0,C+3,k)$ is in $S$, contradicting the choice of $A,B,C$.
Thus, the process alternates between case 3 and some other case, say case 2 for concreteness. At each step, the first coordinate of $s_i$ will then increase, and inductively we see that for $i\ge 0$, the first coordinate of $s_i$ is $x+i$. Since it can't increase without bound, the process must eventually end. If it ends after step $0$, then the dominating vector $s_1$ is either $(x+1,0,0,k+1)$ or $(0,y+1,0,k+1)$. If it ends after step $i$ for $i\ge 1$, then the dominating vector $s_{i+1}$ must be $(x+i+1,0,0,k+1)$, since otherwise it would be either $(0,B+2,0,k+1)$ or $(0,0,C+2,k+1)$ depending on whether $i$ was even or odd, contradicting the choice of $A,B,C$.
Best Answer
I guess the answer should be that $(a,b,c)$ fills the line if and only if there is an integral combination of its permutations equaling $(1,1,1)$. We want the elementary divisors of a certain $3 \times 6$ matrix. Working out explicit conditions should be do-able. I wonder if there is a slick argument.
There is also a neat connection with semi-magic squares, I think.
When allowing longer vectors which are permutations of $(a,b,c,0,\dots,0)$, it feels like their Frobenius number is the truth, but again with some extra necessary divisibility conditions.
(Sorry if my question was elementary.)