Note: Your polynomials are still polynomials in one variable. It's the coefficients of the polynomial that are rational functions in several variables, but these coefficients are constants of your field.
To show that $F(u_1,\ldots,u_n)$ is Galois over $F(s_1,\ldots,s_n)$, it suffices to show that it is the splitting field of some separable polynomial with coefficients in $F(s_1,\ldots,s_n)$. Let $f(t)\in F(s_1,\ldots,s_n)[t]$ be the polynomial
$$f(t) = t^n - s_1t^{n-1} + s_2t^{n-2} + \cdots + (-1)^ns_n.$$
Since
$$f(t) = (t-u_1)(t-u_2)\cdots(t-u_n),$$
it follows that $F(u_1,\ldots,u_n)$ is indeed the splitting field of $f(t)$ over $F(s_1,\ldots,s_n)$.
Note also that $f(t)$ has no repeated roots, since the $u_i$ are pairwise distinct indeterminates, so $f(t)$ is separable. Thus, $F(u_1,\ldots,u_n)$ is the splitting field of a separable polynomial over $F(s_1,\ldots,s_n)$, and so it is a Galois extension of $F(s_1,\ldots, s_n)$.
To finish off, we need to show the Galois group is precisely $S_n$.
Note that $S_n$ acts on $F(u_1,\ldots,u_n)$ by fixing $F$ and letting $\sigma$ map $u_i$ to $u_{\sigma(i)}$; these automorphisms leave $F(s_1,\ldots,s_n)$ fixed, and they are pairwise distinct, so this proves that $G=\mathrm{Gal}(F(u_1,\ldots,u_n)/F(s_1,\ldots,s_n))$ contains a subgroup isomorphic to $S_n$. On the other hand, an element of $G$ is completely determined by what it does to $u_1,\ldots,u_n$, and it must map $u_i$ to a root of $f(t)$ above, hence elements of $G$ act as permutations of $u_1,\ldots,u_n$, showing that $G$ must be isomorphic to a subgroup of $S_n$. These two conclusions tell us that $G$ must in fact be isomorphic to $S_n$, as desired.
Another method is to note, as KCd remarks, that $S_n$ acts on $F(u_1,\ldots,u_n)$, hence, $F(u_1,\ldots,u_n)$ is Galois over the fixed field of $S_n$, with Galois group $S_n$. The fixed field of $S_n$ under this action is precisely the field of symmetric rational functions, which by the Fundamental Therorem of symmetric functions is generated over $F$ by $s_1,\ldots,s_n$; that is, the fixed field is $F(s_1,\ldots,s_n)$, which shows that $F(u_1,\ldots,u_n)/F(s_1,\ldots,s_n)$ is Galois with Galois group $S_n$, as desired.
Best Answer
I don't see how to use Eisenstein here. But the polynomial is a cubic, so if it factors, then one of the factors is linear, and the polynomial would have a zero $x=\frac{p(t)}{q(t)}\in\Bbb{C}(t)$ with some polynomials $p(t),q(t)\in\Bbb{C}[t]$, $q\neq0$.
To exclude this possibility we use a line of reasoning analogous to the familiar rational root test - taking advantage of the fact that $\Bbb{C}[t]$ is a UFD. So assume that all common factors of $p(t)$ and $q(t)$ have been cancelled. Then $$ x^3-x+t=\frac{p^3-pq^2+tq^3}{q^3}=0. $$ Here the numerator has to be zero, so from $$ p^3-pq^2=-tq^3 $$ we can conclude that $p$ divides the left hand side, hence also the right hand side. But $p$ has no common factors with $q$, so it has to be a factor of $t$.
Similarly from $$ p^3=pq^2-tq^3 $$ we see that $q$ divides the right hand side, hence also $p^3$. But, as above, this implies that $q$ must be a constant.
The non-zero constants are the units of $\Bbb{C}[t]$, so we can conclude that $x=p/q$ is either a constant or, $x=at$ for some $a\in\Bbb{C}$.
I'm sure that you can show that neither of those work. Therefore this cubic has no linear factors over $\Bbb{C}(t)$ and hence it is irreducible.