The two notions are not equivalent in the generality that is stated, but one gets an equivalence by making a slight change. The change is that rather than saying that the $\sigma$-algebra is countably generated, what you get will be that it is contained within the completion of a countably generated $\sigma$-algebra.

Note that the second definition you cite is only defined when there is also a measure $\mu$ on the $\sigma$-algebra, and the distance between two sets in the $\sigma$-algebra is defined to be the measure of their symmetric difference $d(A,B)=\mu(A\triangle B)$.
 
For a counterexample to the  strict interpretation of your question, let $X$ be any uncountable set and $S$ the full power set of $X$. This is not countably generated as a $\sigma$-algebra. But fix any $p\in X$ and let $\mu$ be the measure placing mass $1$ at $p$, with no measure outside of {p}. In this case, the family {emptyset, {p}} is dense in your semi-metric, since every subset is essentially empty or {p}. So the semi-metric is separable, but the $\sigma$-algebra is not countably generated. QED (More interesting examples can easily be constructed without point masses using the same idea: add an uncountable set of measure $0$ and use the completion of the $\sigma$-algebra.) 

For the positive result, note that the counterexample $\sigma$ algebra was the completion of the finite Boolean algebra generated by {p}. So you should really be speaking of the $\sigma$-algebra being (contained in) the completion of a countably generated $\sigma$-algebra. And with this version of the question, you get the desired equivalence. 

To prove the desired equivalence, suppose that a (finite) measure is defined on a $\sigma$-algebra $S$ that is generated by a countable subfamily $A_0$, $A_1$, $A_2$, and so on. Every set $A\in S$ is obtained by a union of the form $A=\bigcup_{n\in I} \pm A_n$, where $+A=A$ and $-A$ is the complement of $A$ and $I$ is any set of natural numbers. Let $B_k=\bigcup_{n\in I, n\lt k} \pm A_n$ be the finite approximations, so that  $A=\bigcup_k B_{k+1}-B_k$, which is a disjoint union. Thus, the measure of $B_{k+1}-B_k$ must tend to $0$, and so we can find some $B_k$ as close to $A$ as desired. Thus, the finite Boolean combinations of the $A_n$ are dense in your semi-metric space. 

Conversely, suppose that some countable family $S_0$ is dense in your semi-metric space. Let $S$ be the $\sigma$-algebra generated by these sets. For any $A\in S$, we may find $A_n\subset A$ with $A_n\in S_0$ and $d(A,A_n)\lt \frac1n$. Thus, $\bigcup_n A_n$ is in $S$, and differs from $A$ on a set of measure $0$. So $A$ is in the completion of $S$, as desired.