We can characterize Banach limits as continuous functionals on $\ell^\infty$ which vanish on
$$ X := \{(x_n - x_{n+1}): (x_n) \in \ell^\infty\} $$
and which send the constant sequence $(1,1,\dots)$ to $1$.

Note that $X$ is a subspace.
The Hahn-Banach Theorem tells us that we are asking: if $(y_n) \in \ell^\infty$ has Cesaro mean $0$, is it in the closure of $X$?
(And the converse question is: does every element of $X$ have Cesaro mean $0$?
Yes; since the $n^\text{th}$ Cesaro mean of $(x_n-x_{n+1})$ is $(x_1-x_{n+1})/n$, which converges to $0$ since $(x_n)$ is uniformly bounded.)

The answer is no.
Consider the sequence $(y_n)$ that has $1$ once, followed by $-1$ three times, then $1$ five times, and so on.
One can compute the Cesaro mean, and see that it approaches $0$ in the limit.
But $(y_n)$ is not in the closure of $X$.

Surely, if it were, then let $(x_n) \in \ell^\infty$ be such that
$$ \|(y_n) - (x_n-x_{n+1})\|_\infty < 1/2. $$
Let $M$ be a natural number, $M \geq \|(x_n)\|$.
Let $n$ be an index such that
$$ y_n = \cdots = y_{n+4M} = 1. $$
Then for $i=1,\dots,4M$,
$$ x_{n+i} < x_{n + i-1} - y_{n + i - 1} + 1/2 = x_{n + i - 1} - 1/2, $$
and summing these up, we find
$$ x_{n+4M} < x_n - 4M/2. $$
This contradicts the assumption that $\|(x_n)\| \leq M$.