That seems an unnecessary approach here. First prove the theorem for non-negative functions, then extend to all functions as you've indicated.
For nonnegative $f$, note that if $h$ is bounded and has finite support on $A$, then you can define $$h_E(x) =\begin{cases} h(x) & x\in A\\0 & x \in E \setminus A\end{cases}$$
Then $h_E$ is bounded and has finite support on $E$. Further, $h_E\chi_A = h_E$ and $h_E|_A = h$. Conversely if $h$ is bounded and has finite support on $E$, then $h\chi_A|_A$ is bounded and has finite support on $A$.
Edit: as requested, where I've tried to dot all the i's and cross all the t's.
For notational convenience, for measurable sets $B$ and measurable non-negative functions $g$, define ${\scr M}(g, B)$ to be the set of all bounded measurable functions $h$ of finite support on $B$ such that $0 \le h \le g$.
Let $f$ be a non-negative measurable function on $E$ and $A$ a measurable subset of $E$. Then $\chi_Af$ is also a measurable function on $E$.
Now if $h \in {\scr M}(\chi_Af,E)$ and $x \in E \setminus A$, then $0 \le h(x) \le \chi_A(x)f(x) = 0\cdot f(x) = 0$. Hence $\operatorname{supt} h \subseteq A$. And for $x \in A$, $0 \le h(x) \le \chi_A(x)f(x) = 1\cdot f(x) = f(x)$, so $0 \le h \le f$ on $A$. Therefore $h|_A \in {\scr M}(f,A)$. And, $$\int_E h = \int_A h + \int_{E\setminus A} h = \int_A h$$ since $h = 0$ on $E\setminus A$.
Conversely, if $h' \in {\scr M}(f,A)$, then define $h : E \to \bar{\Bbb R}$ by $$h(x) = \begin{cases} h'(x) & x \in A\\0 & x \notin A\end{cases}$$
Then $h$ is measurable, $\operatorname{supt} h = \operatorname{supt} h'$ and therefore is finite, and if $h' < M$, then $M > 0$, so $h < M$ as well. And lastly $$h(x) = \begin{cases} h'(x) \le f(x) = \chi_A(x)f(x) & x \in A\\0 = \chi_A(x)f(x) & x \notin A\end{cases}$$
Hence $h \in {\scr M}(\chi_Af,E), h' = h|_A$ and therefore $$\int_E h = \int_A h'.$$ Therefore the restriction map $T\ :\ {\scr M}(\chi_Af,E) \to {\scr M}(f,A)\ :\ h \to h_A$ is a bijection, and $\int_E h = \int_A T(h)$. So $$\begin{align}\int \chi_Af &= \sup\left\{\int_E h \ :\ h \in {\scr M}(\chi_Af,E)\right\}\\
&=\sup\left\{\int_A T(h) \ :\ h \in {\scr M}(\chi_Af,E)\right\}\\
&=\sup\left\{\int_A T(h) \ :\ T(h) \in {\scr M}(f,A)\right\}\\
&=\sup\left\{\int_A h' \ :\ h' \in {\scr M}(f,A)\right\}\\
&=\int_A f\end{align}$$
When $f$ may be negative, $$\int_E\chi_Af = \int_E\chi_Af^+ -\int_E\chi_Af^- = \int_A f^+ - \int_Af^- = \int_Af$$
Best Answer
Yes, your proof for characteristic functions case is fine.
I guess that you're trying to use the standard machine to extend that result to all bounded measurable functions. It usually goes like this :
You are at step 2, you thus only need to prove the result for non-negative simple functions first, so take all your $a_i$'s to be non-negative.
Edit to address the comments:
As I said, this is just step 2 out of 4 in the standard machine, which will then allow you to extend the property to all measurable functions. If you're not familiar with this approach, it is a very effective and thus very popular way to generalize results which are true for simple functions to general bounded measurable functions.
The idea is that, instead of trying to prove directly a result for all measurable functions, you first prove it for characteristic functions, which are generally much easier to deal with, and from there progressively generalizing to non-negative simple functions, non-negative measurable functions and finally arbitrary measurable functions. You have examples of this method being used to solve a some measure theory problems e.g. here, here or here.
Lastly, here is a quote from Chapter 3 of Prof. Gordan Žitković's lecture notes on probability theory, which essentially restates what I just wrote (some more examples are given in the notes) :
I suggest you read further on the topic to familiarize yourself with the idea.
As you wrote, if $\psi$ is a non-negative simple function, it can be written as $ \psi = \sum_{i=1}^{n} a_i \chi_{E_i}$. And similarly $$ \psi\cdot\chi_{C} = \sum_{i=1}^{n} a_i \chi_{E_i}\cdot\chi_{C} $$ So using the linearity of Lebesgue integral and the fact that you have proven the property for characteristic functions $$\begin{align} \int_E \psi\cdot\chi_C &= \int_E \sum_{i=1}^{n} a_i \chi_{E_i}\cdot\chi_{C}\\ &=\sum_{i=1}^{n} a_i \int_E \chi_{E_i}\cdot\chi_C\tag1\\ &= \sum_{i=1}^{n} a_i \int_C \chi_{E_i}\tag2\\ &= \int_C \sum_{i=1}^{n} a_i \chi_{E_i} = \int_C \psi \tag3\\ \end{align} $$ Where I used the linearity of Lebesgue integral in $(1)$ and $(3)$, and the fact that the property is true for characteristic functions (which you have proven) in $(2)$.