Consider the following two colorings:

(1) Point $(x, y)$ is in color $i$, $1-i-2$ if and only if $x \equiv i \pmod{2}$.

(2) Point $(x, y)$ is in color $i$, $1 - i - 3$ if and only if $x \equiv i \pmod{3}$.

It is clear that if $S$ exists then $2S$ is an integer. Coloring (1) implies that $S$ could be $1$, $2$, $3$ or greater and coloring (2) shows that $S$ could be $3/2$, $3$ or greater. Therefore if $S$ exists then $S \ge 3$.

We prove that for arbitrary coloring there exists a triangle of area $3$ having all its vertices in one and the same color. Note that for some $d \in \{1, 2, 3\}$ there exist points $A = (x, y)$ and $B = (x+d, y)$ of the same color. Indeed, it suffices to consider the points $(0,0), (1,0), (2,0), (3,0)$. If $m \equiv AB$ and $l$ is parallel to $m$ at distance $\frac{6}{d}$ from $m$ then the line $l \parallel Ox$ is either having points of only two colors or we have a triangle of area $3$.

We say that a color $c$ permits distance $a$ in the line $l$ if there exist two points on $l$ having color $c$ and at distance $a$ apart.

If there exists a distance $a \in \{1, 2, 3, 6\}$ that is permitted by the two colors on $l$ then the line $p \parallel l$ at a distance $\frac{6}{a}$ from $l$ is having all its points in the same color. If such a distance does not exist then one of the colors $l$ permits all distances in the set $\{2, 3, 6\}$.

(Indeed, assume that color $c_1$ does not permit distance $1$, and there exists at least one point $P_0$ from $l$ having color $c_1$. Two neighbors of $P_0$ - points $P_{-1}$ and $P_1$ are colored in $c_2$, so $c_2$ permits distance $2$. Therefore $c_1$ does not permit distance $2$ and points $P_{-2}$ and $P_2$ are of color $c_2$. As above $c_2$ does not permit distance $3$. $c_1$ does not permit distance $3$ and $P_{-3}$ and $P_3$ are of color $c_2$. Thus $c_2$ permits all distances in $\{2, 3, 6\}$.)

In both cases there exists a line $p \parallel Ox$ and color $c$ from $p$ that permits all distances in the set $\{2, 3, 6\}$.

Consider the lines $u_1, u_2, u_3$, parallel to $p$ and such that $u_i$, $i = 1, 2, 3$ is at distance $i$ above $p$. It follows from the above that all such lines are having points only in two colors. Finally, consider all nine points of intersection of lines $u_1, u_2, u_3$ with lines $x = 0$, $x = 3$, $x = 6$. It is easy to see that there exists a triangle of area $3$ having all its vertices colored in the same color.