Yes, conditions 1 and 2 are equivalent (as you already noticed).

Let me start by replacing both definitions with their negations, and replace definition 2 with its $L^1$-version.
A and B below are the negations of 1 and 2 correspondingly, slightly rewritten:

A. There exists a sequence of measurable subsets $\emptyset,X\neq A_n\subset X$ such that for every $s\in S$,
$$(*)\quad \frac{\|(\chi_{A_n}-\mu(A_n)\cdot 1)-s(\chi_{A_n}-\mu(A_n)\cdot 1)\|_1}{\|(\chi_{A_n}-\mu(A_n)\cdot 1)\|_1} = \frac{\mu(A_n \Delta sA_n)}{\mu(A_n)(1-\mu(A_n))}\to 0.$$

B. There exists a sequence of vectors $0\neq v_n$ in $L^1_0(X)$ such that for every $s\in S$,
$$(**) \quad \frac{\|v_n-sv_n\|_1}{\|v_n\|_1} \to 0.$$

To see that A is the negation of 1 is straightforward. To see that B is the negation of 2, recall that the map $L^2\ni u\mapsto \text{sgn}(u)\cdot |u|^2\in L^1$ is uniformly continuous equivariant map on the unit sphere. Use it and a projection mod constants. I am not elaborating, as this is well known.
From now on we work in $L^1$ exclusively, thus $\|\cdot\|=\|\cdot\|_1$.

Clearly A $\Rightarrow$ B. The rest of this post is devoted to proving the application B $\Rightarrow$ A. Thus, we assume that $v_n$ is a sequence of unit vectors satisfying $(**)$ and argue to provide a sequence $A_n$ satisfying $(*)$.

Let me denote by $B$ the unit ball in $L^1_0(X)$.
The idea, as usual, is to map $B$ continuously into a compact space and take a limit point of the image of the sequence $v_n$.
A first candidate for such a map that comes to mind is the embedding into the unit ball of $L^1(X)^{**}\simeq L^\infty(X)^*$. This idea will lead us eventually to prove the existence of an invariant mean, as in the answer by Mateusz Wasilewski, but justifying everything down this road seems longer than the path we take below in which we use another map.

Consider the space of probability measures on $\mathbb{R}$, $P$.
Endow $P$ with the weakest topology for which integration against every function in $C_c(\mathbb{R})$ is continuous. Observe that the map $B\to P$, $f\mapsto f_*\mu$ is continuous and that the image is precompact (use Prokhorov's theorem).
Let me denote $\nu_n=(v_n)_*\mu\in P$ and assume as we may that $\nu_n\to \nu$.
We consider two different cases: either $\nu$ is a point mass or not.

In case $\nu$ is *NOT* a point mass we can find and fix $t\in \mathbb{R}$ such that $\nu(-\infty,t),\nu(t,\infty)>0$ and then,
defining $A_n=v_n^{-1}(t,\infty)$, it easy to check that $(*)$ is satisfied (use the continuity of $B\to P$).

From now on we assume $\nu$ is a point mass, that is $\nu=\delta_t$ for some $t\in \mathbb{R}$.
By the fact that $\int v_n=0$ and $\int|v_n|=1$ it is easy to conclude that $|t|\leq 1/2$ and it follows that $1/2\leq \|v_n-t\|<3/2$.
It is also easy to see that the sequence $|v_n-t|$ is almost invariant in $L^1(X)$ and that $|v_n-t|_*\mu \to \delta_0$.
We normalize this sequence, setting $u_n=|v_n-t|/\|v_n-t\|$.

To summarize: we found a new almost invariant sequence of positive unit vectors $u_n\in L^1(X)$ such that $(u_n)_*\mu\to \delta_0$.
We assume as we may that $\sum_{s\in S}\|su_n-u_n\|< \|u_n\|/n$.
The construction of the sequence $A_n$ will follow by the "layer cake decomposition" method: for every positive function $f\in L^1(X)$,
$\int_X f=\int_0^\infty f_*\mu(t,\infty)dt$.
Applying this decomposition to the functions
$|su_n-u_n|$ and $u_n$ we obtain
$$ \sum_{s\in S}\int_0^\infty |su_n-u_n|_*\mu(t,\infty)dt=\sum_{s\in S}\int_X |su_n-u_n|=\sum_{s\in S}\|su_n-u_n\| $$
$$ < \|u_n\|/n = 1/n \int_X u_n =1/n \int_0^\infty (u_n)_*\mu(t,\infty)dt $$
and deduce the existence of $t>0$ for which
$$ \sum_{s\in S} |su_n-u_n|_*\mu(t,\infty) < 1/n (u_n)_*\mu(t,\infty). $$
We set $A_n=u_n^{-1}(t,\infty)$ and conclude
$\sum_{s\in S} \mu(A_n \Delta sA_n) < 1/n \mu(A_n)$.
By the strict inequality we get $A_n\neq \emptyset$.
Since $(u_n)_*\mu\to \delta_0$ we have $1-\mu(A_n)\to 1$, thus we may assume $A_n\neq X$, and indeed we get
$$ \frac{\mu(A_n \Delta sA_n)}{\mu(A_n)(1-\mu(A_n))} \to 0. $$