The topological dual of $\ell^\infty(\mathbb{N})$ is too large to assume that such a functional $\Lambda$ must have a representation as some sequence (see here for a characterisation of the dual). In particular, it is not the case that $\Lambda(Sx) = \Lambda(x)$ implies that $\Lambda$ is the action of a constant sequence.
In what follows, where I use $1$ in the place a sequence should be I mean $(1,1, 1,\dots) \in \ell^\infty$. One approach is the following:
Consider the operator $T = I-S$. Then $\Lambda(Sx) = \Lambda(x)$ for every $x$ iff $T(\ell^\infty) \subset \ker \Lambda$. One can show that the kernel of $T$ consists of the constant sequences and that if $Y = \ker T \oplus T(\ell^\infty)$ then the projection $P$ onto $\ker T$ along $T(\ell^\infty)$ is bounded (with norm $1$).
Now define $\pi_1(x) = x_1$ for $x \in \ell^\infty$ and define $\Lambda_0: Y \to \mathbb{R}$ by $\Lambda_0 = \pi_1 \circ P$. It is easily shown that $\Lambda_0$ is a norm $1$ functional such that $\Lambda_0(1) = 1$.
By Hahn-Banach, there is a norm $1$ functional, $\Lambda$, on $\ell^\infty$ extending $\Lambda_0$. By construction, $T(\ell^\infty) \subset \ker P \subset \ker \Lambda_0 \subset \ker \Lambda$ so $\Lambda(Sx) = \Lambda(x)$. It is less clear that $\Lambda$ has the other two properties. This will follow from the following result which was in fact the rest of my motivation for defining $\Lambda_0$ in the way I did.
Lemma: Suppose $f \in (\ell^\infty)^*$ with $\|f\| = 1$. Then $f(Sx) = f(x)$ for all $x$ and $f(1) = 1$ implies $\lim \inf \{x_n\} \leq f(x) \leq \lim \sup \{x_n \}$ for every $x \in \ell^\infty$.
You can prove this by showing that such an $f$ is then non-negative in the sense that $x_n \geq 0$ for all $n$ implies $f(x) \geq 0$ and then e.g. considering $f(S^nx - \alpha 1)$ for $\alpha < \lim \inf \{x_n\}$ where $n$ is chosen so that $x_{n+m} > \alpha$ for every $m$.
The other two desired properties of $\Lambda$ then follow in a straightforward manner.
In fact, there is another more general technique for proving this result that needs some heavier machinery. As noted in the comments, on the space of convergent sequences $c \subset \ell^\infty$, the limit functional matches the requirements. So one might attempt to take a Hahn-Banach extension $\Phi$ of $\lim : c \to \mathbb{R}$ and hope that this still has the desired properties.
By my Lemma and its definition, $\Phi$ will satisfy the first two properties. However, it is not obvious that $\Phi(Sx) = \Phi(x)$ for every $x$ (this is the reason that I constructed $\Lambda_0$ on a space other than $c$ above; I was most worried about this property).
It is in fact true that you can find a Hahn-Banach extension with the desired property. The key is to recognise that $\Phi(Sx) = \Phi(x)$ for every $x$ iff $\Phi$ is a fixed point of $S^*$. In fact, we actually don't even need to be working with the shift operator $S$ on $\ell^\infty$. More generally, we have
Theorem: Let $Y$ be a subspace of a Banach space $X$ and let $T:X \to X$ be a bounded linear operator. If $g \in Y^*$ and $g(Tx) = g(x)$ for every $x \in Y$ then there is $f \in X^*$ such that $f$ extends $g$, $\|f\| = \|g\|$ and $f(Tx) = f(x)$ for all $x \in X$.
The idea of the proof is to apply the Markov-Kakutani Fixed Point Theorem to $T^*$ defined on the set $\{ f \in X^*: \|f\| = \|g\| \text{ and } f \text{ extends } g \}$. This set is non-empty by Hahn-Banach, is easily seen to be convex and, by an application of Banach-Alaoglu, is weak$^*$-compact since it is clearly weak$^*$ closed. So by Markov-Kakutani, we get the desired functional.
In fact, one can generalise this even further to find Hahn-Banach extensions that preserve the property $f(T_ix) = f(x)$ for every $x$ as long as the $T_i$ commute (again just Markov-Kakutani) but this answer is already quite long so I won't go into details here.
