This argument avoids the use of ultrafilters.
Let $e_n \in \ell_\infty^*$ be the evaluation map $e_n(x) = x_n$. The set $\{e_n\}$ is contained in the unit ball of $\ell_\infty^*$, which by Alaoglu's theorem is weak-* compact, hence $\{e_n\}$ has a weak-* cluster point; call it $f$. I claim this $f$ has the properties you desire.
It is easy to check that the set of multiplicative linear functionals is weak-* closed in $\ell_\infty^*$; each $e_n$ is multiplicative and hence so is $f$.
For each $x \in \ell^\infty$, let $\pi_x : \ell_\infty^* \to \mathbb{C}$ be the evaluation functional $\pi_x(g) = g(x)$. By definition of the weak-* topology, $\pi_x$ is weak-* continuous. Suppose $x \in c \subset \ell_\infty$ is convergent, with $x_n \to a$. By continuity of $\pi_x$, $f(x) = \pi_x(f)$ must be a cluster point of $\{\pi_x(e_n)\} = \{x_n\}$. But $x$ is a convergent sequence so the only cluster point of $\{x_n\}$ is $a$. Thus $f(x) = a$.
$f$ has another interesting property: since $\mathbb{C}$ is metric, all cluster points in $\mathbb{C}$ are subsequential limits. Thus for any $x \in \ell^\infty$, $f(x)$ is a subsequential limit of $\{x_n\}$. For instance, if $x_n = (-1)^n$, $f(x)$ must be either -1 or 1, whereas a Banach limit must assign $x$ the limit value of 0. A corollary of this is that for any real sequence $\{x_n\}$, we must have $\liminf x_n \le f(x) \le \limsup x_n$.
It is also interesting to note that $f$ is an element of $\ell_\infty^*$ that cannot correspond to an element of $\ell^1$. So this gives an alternate proof that $\ell^1$ is not reflexive.
If you like, you can instead produce $f$ as a limit of a subnet of $\{e_n\}$, or a limit of $\{e_n\}$ along an ultrafilter. In any case it is a cluster point.
Pick an $f_{0}\in L^{\infty}-C[0,1]$ and let $W=\text{span}(\{f_{0}\}\cup C[0,1])$. Consider $\Gamma(\alpha f_{0}+g)=\alpha d(f_{0},C[0,1])$ for any scalar $\alpha$, then $\Gamma$ vanishes on $C[0,1]$, certainly $\Gamma$ is a non-zero bounded linear functional on $W$.
Hahn-Banach Theorem gives an extension of $\Gamma$ to the whole $L^{\infty}$.
Best Answer
I guess $S$ is a quotient of $\mathcal{B}$, rather than a subspace, correct?
That is, we may define a projection mapping $\pi :\mathcal{B}\to S, $ by letting $\pi (f)$ be the restriction of $f$ to $\mathbb N$, for every $f$ in $\mathcal B$.
Choosing a Banach limit $\Lambda :S\to \mathbb R$, we then define $\mathcal{L}$ to be the composition $$ \mathcal B \quad {\buildrel \pi \over \longrightarrow} \quad S \quad {\buildrel \Lambda \over \longrightarrow} \quad \mathbb R. $$ The OP has already observed that this functional has all of the required properties, except possibly for translation invariance.
Denoting by $T_y(f)$ the translation of $f$ by $y$, namely $$ T_y(f)|_x = f(x-y), $$ the missing property is thus $$ \mathcal L(T_y(f)) = \mathcal L(f). \tag 1 $$ Observe however that $\Lambda $ is invariant by integer translations, so this easily implies that (1) holds provided $y$ is an integer number. In particular the correspondence $$ y\mapsto \mathcal L(T_y(f)) \tag 2 $$ is seen to be periodic with period 1.
When desired invariance properties are absent, it is a common trick in the theory of group representations to average out. That is, in order to overcome the above difficulty let us define $\tilde{\mathcal L}$ by $$ \tilde{\mathcal L}(f) = \int_0^1 \mathcal L(T_y(f))\, dy. $$ Given any $z$ in $\mathbb R$, we then have that $$ \tilde{\mathcal L}(T_z(f)) = \int_0^1 \mathcal L(T_{y+z}(f))\, dy = \int_z^{z+1} \mathcal L(T_{y}(f))\, dy = \int_0^1 \mathcal L(T_{y}(f))\, dy = \tilde{\mathcal L}(f), $$ where the penultimate step is due to the periodicity of (2).
It is now easy to see that $\tilde{\mathcal L}$ satisfies all of the required conditions.