Let the roots be $a,b,c,d,e$ with $a,b $ non-real. Your subgroup is transitive and has at least one transposition $(a b)$. Since it is transitive you can conjugate your transposition by some element to get a transposition which moves $c$ (conjugation preserves the cycle structure), and likewise one which moves $d$ or $e$.
Case 1. If this is $(bc)$ or $(ac)$ you can find your 3-cycle as a product of transpositions.
Case 2. If it is $(cd)$ then you can use conjugation to find a transposition which moves $e$ and whichever one you get, you can find your 3-cycle by composing it with either $(ab)$ or $(cd)$.
I seem to be giving lots of answers that depend on very special properties of the supplied example. Here’s an argument, tailored to your polynomial $f(x)=x^3-x-1$. Not the general method you were hoping for at all.
First set $\alpha$ to ba a root of your polynomial, which we all know is irreducible over $\Bbb Q$. It’s not too hard to calculate the discriminant of the ring $\Bbb Z[\alpha]$ as the norm down to $\Bbb Q$ of $f'(\alpha)=3\alpha^2-1$; this number is $23$, surprisingly small for a cubic extension. The fact that it’s square-free implies that $\Bbb Z[\alpha]$ is the ring of integers in the field $k=\Bbb Q(\alpha)$.
Our field $k$ clearly is not totally real, since $f$ has only one real root. So in the jargon of algebraic number theory, $r_1=r_2=1$, one real and one (pair of) complex embedding(s). We can apply the Minkowski Bound
$$
M_k=\sqrt{|\Delta_k|}\left(\frac4\pi\right)^{r_2}\frac{n!}{n^n}\,,
$$
which for $n=3$ gives a bound less than $2$, so that $\Bbb Z[\alpha]$ is automatically a principal ideal domain.
Let’s factor the number $23$ there: we certainly know that it’s not prime, since $23$ is ramified.
Now, we already know a number of norm $23$, necessarily a prime divisor of the integer $23$, it’s $3\alpha^2-1$, and indeed $23/(3\alpha^2-1)=4 + 9\alpha - 6\alpha^2$. But better than that, $23/(3\alpha^2-1)^2=3\alpha^2-4$. This number has norm $23$ (because the norm of $23$ itself is $23^3$). So we’ve found the complete factorization of $23$.
Now let’s look more closely at $f(x)=(x-\alpha)g(x)$, for a polynomial $g$ that we can discover by Euclidean Division to be $g(x)=x^2+\alpha x+\alpha^2-1$. And the roots of $g$ are the other roots of $f$; the Quadratic Formula tells you what they are, and the discriminant of $g$ is $\alpha^2-4(\alpha^2-1)=4-3\alpha^2$. which we already know as $-23/(3\alpha^2-1)^2$. Going back to the Quadratic Formula, our other roots are
$$
\rho,\rho'=\frac{\alpha\pm\sqrt\delta}2\>,\>\delta=\frac{-23}{(3\alpha^2-1)^2}\>,\>\sqrt\delta=\frac{\sqrt{-23}}{3\alpha^2-1}\>.
$$
And that gives you your roots of this one very special cubic polynomial in terms of one root $\alpha$ and $\sqrt{-23}$.
Best Answer
No.
Suppose $f(x) = x^3 + ax + b$ (a linear change of variable will take it to this form). If $\alpha_1,\alpha_2,\alpha_3$ are the three roots of $f(x)$, then $$\delta = (\alpha_1-\alpha_2)(\alpha_1-\alpha_3)(\alpha_2-\alpha_3)$$ lies in the splitting field, and its square is the discriminant of the polynomial, $\delta^2 = -4a^3 - 27b^2$.
So the splitting field contains $\mathbb{Q}(\sqrt{\Delta})$, hence the splitting field has degree $6$ over $\mathbb{Q}$ whenever the discriminant is not a rational square. In fact, the converse also holds: if the splitting field has degree $6$, then the discriminant is not a square.
On the other hand, the splitting field is real if and only if the discriminant is positive (since it equals a square).
So to find an irreducible cubic whose splitting field is completely real but has Galois group $S_3$, you just need to find an irreducible cubic whose discriminant is positive and not a square. For example, $f(x) = x^3 -4x + 1$; the only possible rational roots are $1$ and $-1$, so it is irreducible, and the discriminant is $$\Delta = -4(-4)^3 -27 = 229,$$ which is positive and not a square. So the splitting field is of degree $6$ and has Galois group $S_3$.