The two notions are not equivalent. Indeed, they are not
equivalent even when one considers completing the measure
by adding all null sets with respect to any countably generated $\sigma$-algebra.

The easiest counterexample for the strict question asked is
to let $X$ be an uncountable set and $S=P(X)$, the
full power set of $X$. This is a $\sigma$-algebra, but it
is easily seen not to be countably generated. Fix any $p\in
X$ and let $\mu$ be the measure placing mass $1$ at $p$ and
0 mass outside {p}. In this case, the family {emptyset, X}
is dense in the semi-metric, since every subset is
essentially empty or all of $X$, depending on whether it
contains $p$. So the semi-metric is separable, but the
$\sigma$-algebra is not countably generated.

Note that in this counterexample, the $\sigma$ algebra is
obtained from the counting measure on {p} by adding an
uncountable set of measure $0$ and taking the completion.
Similar counterexamples can be obtained by adding an
uncountable set of measure $0$ to any space and taking the
completion.

At first, I thought incorrectly that one could address the issue by considering the completion of the measure, and showing that the $\sigma$-algebra would be contained within the completion of a countably generated $\sigma$-algebra. But I now realize that this is incorrect, and I can provide a counterexample even to this form of the equivalence.

To see this, consider the filter
$F$ of all sets $A\subset \omega_1$ that contain a closed
unbounded set of countable ordinals. This is known as the
*club filter*, and it is closed under countable
intersection. The corresponding ideal $NS$ consists of the
*non-stationary sets*, those that omit a club, and these
are closed under countable union. It follows that the
collection $S=F\cup NS$, which are the sets measured by a
club set, forms a $\sigma$-algebra. The natural measure
$\mu$ on $S$ gives every set in $F$ measure $1$ and every
set in $NS$ measure $0$. This is a countably additive
2-valued measure on $S$. Note that every set in $S$ has
measure $0$ or $1$; in particular, there are no disjoint
positive measure sets. It follows that the family
{emptyset,$\omega_1$} is dense in the semi-metric, since
every set in $S$ either contains or omits a club set, and
hence either agrees with emptyset or with the whole set on
a club. Thus, the semi-metric is separable. But for any
countable subfamily $S_0\subset S$, we may intersect the clubs
used to decide the members of $S_0$, and find a single club
set $C\subset\omega_1$ that decides every member of $S_0$, in
the sense that every member of $S_0$ either contains or
omits $C$. This feature is preserved under complements,
countable unions and intersections, and therefore $C$
decides every member of the $\sigma$-algebra generated by
$S_0$. The completion of the measure on the
$\sigma$-algebra generated by $S_0$ is therefore contained
within the principal filter generated by $C$ together with
its dual ideal. This is not all of $S$, since there are
club sets properly contained within $C$, such as the set of
limit points of $C$. Thus, this is a measure space that has a separable
semi-metric, but the $\sigma$-algebra is not contained in
the completion of any countably generated $\sigma$-algebra.