Okay, I think I finally get the question. So let's say $X$ has $\mu$ distribution which is absolutelly continuous with density $g$ and CDF $G$. Let's say that $\delta_a$ is delta dirac distribution, that is $\delta_a(A) = \begin{cases} 1 & a \in A \\ 0 & a \not \in A \end{cases}$. Moreover, when we know the value of $X$, then random variable $S$ is distributed $\nu(\cdot |X)$, where $\nu(A|x) = \alpha \mu(A) + (1-\alpha)\delta_x(A)$.
So look at $S$ only and try to find it's distribution: $\mathbb P(S \in A) = \mathbb E[\mathbb P(S \in A | X)] = \mathbb E[ \nu(A|X)] = \alpha \mu(A) + (1-\alpha) \mathbb E[\delta_X(A)] = \alpha \mu(A) + (1-\alpha)\mathbb P(X \in A) = \mu(A)$, so $S$ is distributed exactly the same as $X$.
Moreover, looking at joint behaviour: $\mathbb P( (X,S) \in A \times B ) = \mathbb E[ \mathbb P(X,S) \in A \times B | X) ] = \mathbb E [\mathbb E[ 1_{ \{(X,S) \in A \times B \}} | X ]] = \mathbb E[ 1_{ \{X \in A\}} \mathbb E[ 1_{\{S \in B\}} | X]] = \mathbb E[1_{\{X \in A\}} \nu(B|X)] = \alpha\mu(B)\mathbb E[1_{\{X \in A\}}] + (1-\alpha)\mathbb E[1_{\{X \in A\}} 1_{\{ X \in B \}}]= \alpha \mu(A)\mu(B) + (1-\alpha) \mu(A \cap B)$
We can guess the conditional distribution of $X$ given $S$. The guess is:
$\mathbb P(X \in A |S) = \alpha \mu(A) + (1-\alpha)1_{\{S \in A\}}$
Why? Note it is $S$ measurable random variable, moreover for any $C = \{S \in B\}$ we get:
$\mathbb E[(\alpha \mu(A) + (1-\alpha) 1_{\{S \in A\}} ) 1_{\{ S \in B \}}] = \alpha \mu(A) \mathbb P(S \in B) + (1-\alpha) \mathbb P( S \in A \cap B) $ so using the fact that $S$ has the $\mu$ distribution, too, we get the result:
given $S$, random variable $X$ is distributed as $\lambda(\cdot | S)$, where $\lambda(A|s) = \alpha \mu(A) + (1-\alpha) \delta_s(A)$
Now: $$\mathbb E[X | S ] = \int x d\lambda(x|S) = \int x d(\alpha \mu(x) + (1-\alpha) \delta_S(x)) = \alpha \int x d\mu(x) + (1-\alpha) \int x d\delta_S(x) $$ Since $$ E[X] = \int x d\mu(x)$$ and $$ \int_x d\delta_S(x) = S$$ then our result is:
$$ \mathbb E[X|S] = \alpha \mathbb E[X] + (1-\alpha)S$$
I am sure I am misinterpreting something, could someone indicate me what ?
No, your confusion is because that answer has some errors.
Firstly $A$ is neither an event nor a random variable, so you cannot condition on it alone. It is the domain for $Y$ where $X=1$. (So the events $X=1$ and $Y\in A$ are identical).
Secondly, the integration over the domain of $A$ is not a conditional expectation.
$$\int_A y~f_Y(y)~\mathrm d y=\mathsf E(Y;Y\in A) = \mathsf E(Y\mathbf 1_{Y\in A})$$
In short, that answer should be showing that: $$\mathsf E(Y\mid X=1)=\mathsf E(Y~\mathbf 1_{X=1})/\mathsf P(X=1)$$
Best Answer
The expression for $f_{X\mid A}$ is correct.
But $E(X\mid A)$ is not the same as $E(1_AX)$. To understand why, first think about the expression $E(X\mid A)$. When $a$ tends to $b$ we expect it to lay in the interval $(a,b)$ and thus to be roughly equal to $a$ and $b$. On the other hand the $E(1_AX)$ will go to 0 in that case because (intuitively) the random variable $1_AX$ will be almost everywhere equal to zero.
The rest of your approach is correct and in particular the result that $E(X\mid A)=\frac{E(1_AX)}{\Pr(A)}$ is correct too.