Here is a counter-example showing that at least when $\log m \ll n$ isolation is not possible: with high probability, every weight has exponentially many sets summing to it, so no particular set is "isolated" by its weight. (The example still leaves open the possibility that, say, the minimum-weight set could be likely to be isolated in the case that, say, $m \gg 2^n$.)
Take the set family to be all $2^n$ subsets of $[n]$. Take the weights to be in $\mathbb Z_m$ with addition mod $m$. (Typically one would take, say, $m=n^2$.) Let the element weights $w(x)$ $(x\in [n])$ be selected independently and uniformly at random from $\mathbb Z_m$. Recall that the weight of a given set $S\subseteq [n]$ is defined to be $w(S) = (\sum_{x\in S} w(x))\bmod m$.
Theorem 1. For every $k\ge 0$, with probability at least $1 - 1/2^k$, every weight $x$ in $\mathbb Z_m$
has at least $2^{n - O(k\log m)}$ sets $S$ with weight $w(S)=x$.
(No doubt the probability bound can be improved.)
Proof. For $t\ge 0$ and $x\in\mathbb Z_m$, define $C(t, x) = |\{S\subseteq [t] : w(S) = x\}|$ to be the number of weight-$x$ subsets of the first $t$ elements.
So $C(0,0)=1$ (for the empty set), $C(0, x) = 0$ for $x\ne 0$,
and for $t\ge 1$ and $x\in\mathbb Z_m$
$$C(t, x) = C(t-1, x) + C(t-1, x -_m w(t)),\hspace{1in}(1)\hspace{-1in}$$
where $-_m$ denotes subtraction mod $m$.
Define $m_t = |\{x\in\mathbb Z_m : C(t, x) \ge 1\}|$ to be the number of distinct set weights achieved by the end of round $t$. Let r.v. $T$ be the number of rounds until $m_T = m$, that is, $C(t, x)\ge 1$ for all $x\in\mathbb Z_m$.
(Technically, in this definition we are imagining that the random experiment goes on indefinitely---more than $n$ rounds---adding a new random weight each time.)
Lemma 1. $E[T] \le 1+3\ln m$
Proof of Lemma 1. By (1), in each round $t \le T$, any given weight $x\not\in W_{t-1}$ enters $W_t$ if $x\in w(t) +_m W_{t-1}$, that is, $x -_m w(t) \in W_{t-1}$,
which (given that $w(t)$ is random, so $x -_m w(t)$ is random, and $|W_{t-1}|=m_{t-1}$) happens with probability $m_{t-1}/m$. It follows that
$$\textstyle
E[m_t - m_{t-1} \,|\, m_{t-1}=i] = (m-m_{t-1})\frac{m_{t-1}}{m} = \frac{1}{1/m_{t-1} + 1/(m-m_{t-1})}.\hspace{.6in}(2)\hspace{-.6in}
$$
For $j\le m$ define
$\textstyle F(j) = \sum_{i=1}^j \frac{2}{i} + \frac{1}{\max(m-i, 1)}.$
Using the definition of $F$ and $m_{t-1} \le m_t \le 2 m_{t-1}$, we have $F(m_t)-F(m_{t-1})$ is at most
$$\textstyle
\sum_{i=m_{t-1}+1}^{m_{t}} \frac{2}{i} + \frac{1}{\max(m-i, 1)}
\ge
(m_{t} - m_{t-1})(\frac{2}{m_{t}} + \frac{1}{m-m_{t-1}}
\ge
(m_{t} - m_{t-1})(\frac{1}{m_{t-1}} + \frac{1}{m-m_{t-1}}),
$$
which with (2) implies $E[F(m_t)-F(m_{t-1}) \,|\, m_{t-1}] \ge 1$.
By Wald's equation, $$E[F(m_T)] \ge F(m_0) + E[T] = E[T].$$
This implies $E[T] \le E[F(m_T)] = F(m)$.
By calculation $F(m) = 2 H_m + H_{m-1} + 1 \le 1+3\ln m$.$~~~~\Box$
Let $B = 2+6\ln m$.
By Lemma 1 and the Markov bound, the probability of the "bad" event $T \ge B$ is at most $1/2$.
So, with probability at least $1/2$, the sets $S\subseteq [B]$ generate all $m$ weights.
By applying the same argument to the sets $S\subseteq \{B+1,B+2,\ldots, 2B\}$, with probability at least $1/2$, those sets generate all $m$ weights.
Likewise, partitioning the first $k B$ indices (in $[kB]$) into $k$ groups of size $B$,
each group of sets generates all weights with probability at least $1/2$,
so (using independence of the groups) the probability of the bad event that none of the group's sets generates all the weights is at most $1/2^k$.
Assume this bad event does not happen.
Then, at the start of round $t_0=kB$,
we have $\min_x C(t, x) \ge 1 = 2^0$ for all $x$.
Then at the end of each round $t\ge t_0$, recalling (1), we have $$\min_x C(t, x) \ge 2\min_{x'} C(t-1, x'),$$
so inductively $\min_x C(t, x) \ge 2^{t-t_0}$.
So, by the final round $t=n$ we have
$$\min_x C(n, x) \ge 2^{n-t_0} = 2^{n-kB} = 2^{n-O(k \log m)}.~~~~~\Box$$
