= Solution
Let $r$ be the number at risk just after $a$, and suppose the $d$ events are grouped into tie sizes $d_1,\ldots,d_m$ at ordered times in $(a,b]$. With no intervening censoring or entry, their risk counts are $r$, $r-d_1$, $r-d_1-d_2$, and so on. Hence the <Kaplan–Meier telescoping across a censor-free interval> gives
$$
\frac{\widehat F(b)}{\widehat F(a)}
=\prod_{j=1}^m
\frac{r-\sum_{k\leq j}d_k}{r-\sum_{k<j}d_k}
=\frac{r-d}{r},
\qquad
\boxed{\widehat F(b)=\widehat F(a)\frac{r-d}{r}.}
$$
Only the total $d$ appears. Consequently \b[the order of the events and their grouping into ties do not affect the estimate at $b$]. Censoring exactly at $b$ does not change this conclusion when it is processed after the events at that time, as in the standard convention. The calculation assumes $r>0$; no estimator beyond an exhausted <risk set> is being inferred from unobserved follow-up.
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