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. Nevertheless, the forward
implication holds.

First, let me explain the forward implication. Suppose that
$S$ is a $\sigma$-algebra generated by a countable
subfamily $S_0$ and $\mu$ is a finite measure defined on
$S$. The semi-metric on $S$ is defined by
$d(A,B)=\mu(A\triangle B)$. Let $S_1$ be the collection of
finite Boolean combinations of sets in $S_0$. This is a
countable family, and I claim it is dense in the
semi-metric. To see this, let $S_2$ be the closure of $S_1$
in the semi-metric, that is, the sets $A\in S$ that are
approximable by sets in $S_1$, in the sense that for any
$r\gt 0$ there is $B\in S_1$ such that $d(A,B)\lt r$. Note
that $S_2$ contains $S_1$ and is closed under complement
since the measure was finite. I claim it is also closed
under countable unions: if each $A_n$ is approximable by
$B_n$ to within $r/2^n$, then $\cup_n A_n$ is approximated
by $\cup_n B_n$ to within $r$, and so one may find an
approximating finite union. So $S_2$ is actually a
$\sigma$-algebra, and since it contains $S_0$, it follows
that $S_2=S$. That is, every set in $S$ is approximable by
sets in $S_1$, and so $S_1$ is a countable dense set in the
semi-metric, as desired.

Let's turn now to the reverse implication, which is not
generally true. The easiest counterexample for this in the
strict sense of the question is to let $X$ be a set of size
continuum 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 on cardinality grounds. 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 a large
cardinality set of measure $0$ and taking the completion.
Similar counterexamples can be obtained by adding such
large cardinality 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.