Yes, it is bijective.

p4sch has already shown the map is well-defined and injective (in fact, an isometry). I claim it is also surjective. (I assume throughout that $\mu$ is a positive measure.)

Note it is an exercise to show that $\newcommand{musf}{\mu_{\rm sf}} \musf$ is indeed a countably additive measure.

Lemma. Suppose $E \in \Sigma$ with $\musf(E) < \infty$. Then there exists a measurable $B \subseteq E$ with $\mu(B) = \musf(E)$.

Proof. By definition of $\musf$, for each $k$ we may find a measurable $A \subseteq E$ with $\infty > \mu(A_k) \ge \musf(E) - 1/k$. Set $B_n = A_1 \cup \dots \cup A_n$. Then $B_n \subseteq E$ and $\mu(B_n) \ge \musf(E) - 1/n$. Moreover, since $\mu(B_n) < \infty$, we have $\mu(B_n) \le \musf(E)$ by the definition of $\musf$. Now set $B = \bigcup_{n=1}^\infty B_n$. Continuity from below shows $\mu(B) \le \musf(E)$ and monotonicity shows $\mu(B) \ge \musf(E)$. So $B$ is as desired. $\Box$

In particular we have $1_B \in L^2(\mu)$. As noted by p2sch, whenever $\mu(B) < \infty$ we have $\mu(B) = \musf(B)$. Hence $\musf(B) = \musf(E)$, so that $\musf(E \setminus B) = 0$ and thus $1_B = 1_E$ $\musf$-a.e. Hence $1_E$ is in the image of $i$. By linearity, every $\musf$-simple function is in the image, so the image is dense in $L^2(\musf)$. But we previously showed $i$ is an isometry, so the image is also closed.

Yes, it is bijective.

p4sch has already shown the map is well-defined and injective (in fact, an isometry). I claim it is also surjective. (I assume throughout that $\mu$ is a positive measure.)

Note it is an exercise to show that $\newcommand{musf}{\mu_{\rm sf}} \musf$ is indeed a countably additive measure.

Lemma. Suppose $E \in \Sigma$ with $\musf(E) < \infty$. Then there exists a measurable $B \subseteq E$ with $\mu(B) = \musf(E)$.

Proof. By definition of $\musf$, for each $k$ we may find a measurable $A \subseteq E$ with $\infty > \mu(A_k) \ge \musf(E) - 1/k$. Set $B_n = A_1 \cup \dots \cup A_n$. Then $B_n \subseteq E$ and $\mu(B_n) \ge \musf(E) - 1/n$. Moreover, since $\mu(B_n) < \infty$, we have $\mu(B_n) \le \musf(E)$ by the definition of $\musf$. Now set $B = \bigcup_{n=1}^\infty B_n$. Continuity from below shows $\mu(B) \le \musf(E)$ and monotonicity shows $\mu(B) \ge \musf(E)$. So $B$ is as desired. $\Box$

In particular we have $1_B \in L^2(\mu)$. As noted by p2sch, whenever $\mu(B) < \infty$ we have $\mu(B) = \musf(B)$. Hence $\musf(B) = \musf(E)$, so that $\musf(E \setminus B) = 0$ and thus $1_B = 1_E$ $\musf$-a.e. Hence $1_E$ (or, more properly, its equivalence class) is in the image of $i$. By linearity, every $\musf$-simple function is in the image, so the image is dense in $L^2(\musf)$. But we previously showed $i$ is an isometry, so the image is also closed.