Edit: I missed some things in my first answer and updated my answer accordingly.
Edit 2: Not only that I missed some things in my first answer, I missed the point completely. I leave the answer here for instructional purposes. Tasadduk's answer is definitely the way to go about 2).
I agree with mac that the first solution is almost ok, but some extra care has to be taken (the supremum might be infinite and even if one could argue that $\infty \cdot 0 = 0$ in measure theory, a clear cut argument is certainly preferable to a convention).
I would first prove that for a non-negative and bounded functions $f$ we have $\int_{A} f \,d\mu = 0$, for this you can use your argument. The monotone convergence theorem then implies that it holds for all non-negative measurable functions, simply by approximating $f$ by $f_{n} = \min{\{f,n\}}$ from below.
Remark. The equality $\int_{A} f = 0$ holds true for all measurable $f$, because a general $f$ can be written as $f = f_+ - f_-$ with $f_{+}(x) = \max{\{f(x),0\}}$ and $f_{-}(x) = \max{\{-f(x),0\}}$ and use the definition $\int_{A} f\,d\mu = \int_{A} f_+ \,d\mu - \int_{A} f_-\,d\mu$ provided at least one of the integrals of the right hand side is finite.
As mac also pointed out, there are quite a few problems in the second proof and I don't see how to save it using your idea.
Here's how I would do it. Let $A_{t} = \{|f| \gt t\}$ and note that for $s \lt t$ we have $A_{s} \supset A_t$. Define
$$a_{t} = \int_{\{|f| \gt t\}} |f|\,d\mu = \int_{A_t} |f|\,d\mu.$$
Since $|f| \geq 0$ and $A_{s} \supset A_{t}$ we have $a_{s} \geq a_{t}$ for $s \leq t$. We want to show that $a_{t} \to 0$ as $t \to \infty$. By monotonicity of $t \mapsto a_{t}$, it suffices to show that $a_{n} \to 0$ as $n \to \infty$ runs through the natural numbers. Since $a_{n} \geq 0$ and $a_{n}$ is monotonically decreasing, the limit $a = \lim_{n \to \infty} a_{n}$ exists, and we want to show that $$ a = 0.$$
Now let $f_{n}(x) = \min{\{|f(x)|,n\}}$ and note that $f_{n} \to |f|$ pointwise (and monotonically). By the monotone convergence theorem (or the dominated convergence theorem, applicable since $f_{n} \leq |f|$ and $|f|$ is integrable, if you prefer) we have
$$ \int |f|\,d\mu = \lim_{n \to \infty} \int f_{n}\,d\mu.$$
On the other hand $n \leq |f|$ on $A_{n}$ and $0 \leq f_{n} \leq |f|$ on $\Omega$ imply
$$ \int f_{n}\,d\mu = \int_{A_{n}} n \,d\mu + \int_{\Omega \smallsetminus A_{n}} f_{n}\,d\mu \leq \int_{A_{n}} |f|\, d\mu + \int |f| \,d\mu = a_{n} + \int |f|\,d\mu.$$
Passing to the limit on both sides of the estimate $\int f_{n} \leq a_{n} + \int |f|$ we get
$$\int |f|\,d\mu = a + \int |f|\,d\mu$$
and as $\int |f|\,d\mu \lt \infty$ we conclude $a = 0$, as we wanted.
In addition to the previous advice, note that the function you gave does not "approximate" $f$. An approximating sequence $\{s_n\}$ for $f$ (which $f$ would have iff $f$ were measurable provided the measure space for the domain of $f$ is complete) would need to be within a distance of $\epsilon$ from $f$ for any given $\epsilon >0$. However, the function $s$ you gave as an "approximation" cannot approximate $f$, in the sense that the sequence $\{s_n\}$ with $s_n=s$ $ \forall n \in \mathbb{N}$ gets no closer than $1$ from $\chi_{[0,1]}$.
Again, it is important to note that a function $f$ on a complete measure space $X$ is measurable if and only if $f$ is the pointwise limit of some sequence of simple functions -or, trivially, is a simple function itself. Consequently, any sequence of "simple" functions approximating a nonmeasurable function must contain a "simple" function with the characteristic function of a nonmeasurable set as part of its construction. In a sense, this is why you must include the assumption that $f$ is measurable in your definition for the Lebesgue integral.
To wit, recall that the integral of a characteristic function is the measure of the pullback set in your domain; in the case of your $f$, since $f$ is characteristic, the integral, were it to be defined, is the measure of the pullback, $$\int f d\mu = \mu\{f^{-1}\{1\}\}=\mu\{[0,1]\},$$
but you have not defined the measure for $[0,1]$, which is not even in your $\sigma$-algebra; nor can we infer the measure of $[0,1]$ from the definition you gave for your measure space, as your collection of measurable subsets is already a closed $\sigma$-algebra that is $\sigma$-finite under $\mu$ (hence, your measure space cannot even be extended, in the usual Caratheodory way, to include $[0,1]$ with an accompanying well-defined measurement).
As a curiosity tangential to your question, it is possible for nonmeasurable functions to arise from limits of simple functions in a complete measure space, but such a collection of simple functions must be uncountable. For instance, with Lebesgue measure on $\mathbb{R}$, take $f=\chi_V$ to be the characteristic function of the (uncountable and nonmeasurable) Vitali set $V$ on $[0,1]$, and consider the (uncountable) collection of measurable functions $\{\chi_v\}, v\in V$. Then $\chi_V=\sup \{\chi_v\}_{v\in V}$. Were we to define an integral as you wish, then in this case, you may want to say that the integral $\int \chi_V = \sup \{ \int \chi_v \} = 0$. But, again, since $\chi_V$ is itself characteristic, we should then have $\mu(V)=0$, but $\mu(V)$ is not defined for $V$ under the Lebesgue measure.
To address your request for a resource, see Royden's Real Analysis, 4th ed., chapters 17 and 18 (particularly pp. 362-363 were helpful as a reference to me for this post).
Best Answer
I think you are thinking about this the wrong way. One should adopt the definition:
$$\int_E f d \mu := \int_{\mathbb{R}} f \chi_E d \mu.$$
Because the Lebesgue measure is complete, $f \chi_E$ really is a measurable function, so the right side is well-defined, and provides a suitable definition of the left side. Specifically you do indeed get $\int_E f d \mu = 0$.
An alternative is to make $E$ a measure space in its own right by equipping it with the $\sigma$-algebra $\mathcal{A}$ consisting of elements of $\mathcal{M}$ which are subsets of $E$, and the restriction of $\mu$ to $\mathcal{A}$. Then again because the Lebesgue measure is complete, $\mathcal{A}$ is actually the power set of $E$, and so $f$ is measurable as a function from this new measure space. The end result is the same.