In a regular space, every open set $U$ is a union of regular open sets whose closures are in $U$. And for two regular open sets $U$, $V$, we have $U ∪ V ⊆ W := \operatorname{int}(\overline{U ∪ V}) ⊆ \overline{U} ∪ \overline{V}$. So you may obtain the increasing union at least in regular hereditarily Lindelöf spaces (so separable metrizable is fine).
Claim. A topological space $X$ has no regular opens other than $X$ and $\emptyset$ if and only if there is no pair of non-empty disjoint opens in $X$.
Proof. Left to right, if there are no regular opens other than $X$ and $\emptyset$, then for each non-empty open $U$ we have $\mathrm{cl}(U) = X$, since otherwise $\emptyset \subsetneq U \subseteq \mathrm{int} (\mathrm{cl} (U)) \subsetneq X$. In other words, $\mathrm{int}(X \setminus U) = \emptyset$, so there is no non-empty open $V$ disjoint from $U$. Right to left, if there is no pair of disjoint opens, then for each non-empty open $U$ we have $\mathrm{int}(X \setminus U) = \emptyset$, so $\mathrm{cl}(U) = X$ and $\mathrm{int}(\mathrm{cl}(U)) = X$. Thus $U$ is not regular unless $U = X$.
Non-example. No regular space which is not indiscrete satisfies this condition: if $F$ is a non-empty and non-total closed set, then regularity produces disjoint non-empty opens.
Non-example. No Hausdorff space with at least two points satisfies this condition: if $x$ and $y$ are distinct points, then the Hausdorff axiom produces disjoint non-empty opens.
Example. Each space where the specialization preorder is upward directed satisfies this condition. For example, the Alexandrov topology of an upward directed partially ordered set satisfies this condition, as does the topology on the reals consisting of intervals of the form $(a, +\infty)$ plus the empty and total set.
Best Answer
Let $\tau^\prime$ denote the topology generated by $\mathrm{Ro}(\tau)$. Note that $\tau^\prime$ is coarser than $\tau$.
I claim that $\tau$ and $\tau^\prime$ have the same regular open sets. To begin I will establish the following.
Fact. If $U \subseteq X$ is $\tau^\prime$-open, then $\operatorname{cl}_{\tau} ( U ) = \operatorname{cl}_{\tau^\prime} ( U )$.
proof of Fact. Since $\tau^\prime$ is coarser than $\tau$ we have that $\operatorname{cl}_{\tau} ( U ) \subseteq \operatorname{cl}_{\tau^\prime} ( U )$.
Given $x \in X \setminus \operatorname{cl}_{\tau} ( U )$, then $V := X \setminus \operatorname{cl}_{\tau} ( U )$ is a $\tau$-open neighborhood of $x$ disjoint from $U$, and since $U$ is $\tau$-open it follows that $\operatorname{cl}_{\tau} ( V ) \cap U = \emptyset$, and therefore $\operatorname{Int}_\tau ( \operatorname{cl}_{\tau} ( V ) ) \cap U = \emptyset$. Note that $\operatorname{Int}_\tau ( \operatorname{cl}_{\tau} ( V ) )$ is regular open in $\tau$, and so it is $\tau^\prime$-open, and clearly $x \in \operatorname{Int}_\tau ( \operatorname{cl}_{\tau} ( V ) )$. Therefore $x \notin \operatorname{cl}_{\tau^\prime} ( U )$.
Using de Morgan's laws from this Fact it follows that for any $\tau^\prime$-closed $F \subseteq X$ we have $\operatorname{Int}_{\tau} ( F ) = \operatorname{Int}_{\tau^\prime} ( F )$.
We know prove that every regular open set in $\tau$ is regular open in $\tau^\prime$, and vice versa.
If $U \subseteq X$ is regular open in $\tau$, then $U$ is $\tau^\prime$-open, and by the Fact and its corollary we have $$ U = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau} ( U ) ) = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau^\prime} ( U ) ) = \operatorname{Int}_{\tau^\prime} ( \operatorname{cl}_{\tau^\prime} ( U ) ). $$ Thus $U$ is regular open in $\tau^\prime$.
If $U \subseteq X$ is regular open in $\tau^\prime$, then using the Fact and its corollary we have $$U = \operatorname{Int}_{\tau^\prime} ( \operatorname{cl}_{\tau^\prime} ( U ) ) = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau^\prime} ( U ) ) = \operatorname{Int}_{\tau} ( \operatorname{cl}_{\tau} ( U ) )$$ and so $U$ is regular open in $\tau$.