The relatively free group $F_{var(G)}(x_1,\dots,x_n)$ is isomorphic to the group of all polynomial functions $G^n\to G$, where a function is called polynomial if it can be expressed via the multiplication and inverse of its arguments; the polynomial functions form a group with respect to the pointwise multiplication.

$F_{var(S_3)}(x,y)$ is not $S_3\times S_3\times C_6$ because the latter group is not two-generated (since it maps onto the elementary abelian group of order 8).

The relatively free group $F_{var(G)}(x_1,\dots,x_n)$ is isomorphic to the group of all polynomial functions $G^n\to G$, where a function is called polynomial if it can be expressed via the multiplication and inverses of its arguments; the polynomial functions form a group with respect to the pointwise multiplication.

$F_{var(S_3)}(x,y)$ is not $S_3\times S_3\times C_6$ because the latter group is not two-generated (since it maps onto the elementary abelian group of order 8).

Edit.

Theorem. The group $F$ of polynomial functions $G^n\to G$ is the relatively free group in $var(G)$ of rank $n$. A free basis of $F$ consists of the functions $f_1(x_1,\dots,x_n)=x_1,\dots,f_n(x_1,\dots,x_n)=x_n$.

Proof. Clearly, $F\in var(G)$. Suppose that we have a relation $w(f_1,\dots,f_n)=1$ in $F$. By definition, this means that the the function the fuction $G^n\to G$ sending $(g_1,\dots,g_n)$ to $w(g_1,\dots,g_n)$ is the constant function identitically equal to 1. Thus, $w(x_1,\dots,x_n)=1$ is an identity (law) in the group $G$. This completes the proof.

Example. The rank-one group $F_{var(S_3)}(x)$ consists of the following 6 functions from $S_3$ to $S_3$: $$ x\mapsto 1,\ x\mapsto x,\ x\mapsto x^2,\ x\mapsto x^3,\ x\mapsto x^4,\ x\mapsto x^5. $$ Note that, according to the definition above, we cannot use constants in formulas for polynomial functions; so, for instance, the function $x\mapsto (12)x$ is not polynomial.

Similarly, the $F_{var(S_3)}(x,y)=\{1,x,y,xy,x^2y,\dots\}$ but I do not know how many different polynomyal functions is there and so I do knot know the order of this group (though this is a question of direct calculation).