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.