If $\mu$ is not a positive measure, the statement is not true in general: If $\mu$ is a signed-measure, use the Hahn-decomposition to conclude that $\mu_{sf}(E) = \mu_{sf}^+(E)$, where $\mu^+$ denotes the positive part. Thus, we lose any information about $\mu^-$.

We know that any integrable function is already located on a $\sigma$-finite measurable subset (that follows from the Chebyshev's inequality). Note that for any set $E \in \Sigma$ with finite measure, i.e. $\mu(E) < \infty$, we have $\mu_{sf}(E) = \mu(E)$, just by definition and  monotonicity. Moreover, if $E$ is $\sigma$-finite (i.e. $E= \bigcup_{n=1}^\infty A_n$ with measurable $A_n$ satisfying $\mu(A_n) <\infty$), then also $\mu(E) = \mu_{sf}(E)$. (Proof: Taking $E_m = \bigcup_{n=1}^m A_n$ and using measure continuity, we get $\mu(E) \leq \mu_{sf}(E)$. On the other hand, we have for any measurable $A \subset E$ already $\mu(A) \leq \mu(E)$ by measure-monotonicity.)

That observation implies for any $f \in L^2(\mu)$
$$\int |f|^2 \, \rm{d} \mu = 2 \int_0^\infty x \, \mu(|f| > x) \, \rm{d} x = 2 \int_0^\infty x \, \mu_{sf}(|f| > x) \, \rm{d} x = \int |f|^2 \, \rm{d} \mu_{sf},$$
i.e. the map is well-defined and injective.

However, the map don't have to be surjective. Depending on the $\sigma$-algebra, we may have not enough measurable sets with finite-measure to cover 'large sets' which are not $\sigma$-finite. One really simple example: Let $\Sigma = \{ \emptyset, A, A^c, \Omega\}$, where $A \neq \Omega,\emptyset$, and define
$$\mu(B) = \begin{cases} 0 & \text{if } B = \emptyset \text{ or } B = A^c \\ \infty & \text{else} \end{cases}.$$ 
This is a measure on $(\Omega,\Sigma)$ such that $1_A \notin L^2(\mu)$, because $\mu(A) = \infty$, but $\mu_{sf} =0$ and hence $1_A \in L^2(\mu_{sf})$.