= Solution
Take $I=\{0,1\}$ and the proper <filter on a set> $F=\{I\}$. In the <first-order language> with unary predicates $P,Q$, let both factors be one-element <first-order structures>. Set $P$ true and $Q$ false in the first factor, and reverse these truth values in the second.
The <reduced product> also has one element $a$. Neither $P(a)$ nor $Q(a)$ holds there: their truth sets are $\{0\}$ and $\{1\}$, neither belonging to $F$. Thus
$$
R\not\models P(a)\vee Q(a),\qquad
\{i:\mathcal M_i\models P(a_i)\vee Q(a_i)\}=I\in F.
$$
\b[<Logical disjunction> can therefore break the equivalence.] A union can belong to a <filter on a set> without either summand belonging to it; the corresponding union property does hold for an <ultrafilter>.
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