Why constant symbols in a language?

An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.

For example, "$\mathbb{Z}$ as a group" and "$\mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, which must be preserved by group homomorphisms.

In your example, a homomorphism of $L$-structures $f : \mathfrak{S}_n \to \mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{\varnothing}$-structures would not.

So while "$\mathfrak{S}_n$ as an $L$-structure" and "$\mathfrak{S}_n$ as an $L_{\varnothing}$-structure" have the same underlying set, they are not the same object.

Fun fact: the assignment from "$\mathfrak{S}_n$ as an $L$-structure" to "$\mathfrak{S}_n$ as an $L_{\varnothing}$-structure" is an example of a forgetful functor.

Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.

For example, if we consider Peano Arithmetic then obviously $\mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > \bar n$ for all $n \in \mathbb N$ (where $\bar n$ stands for 1 added $n$ times: $1 + 1 + \ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.