Is there a clean way to show the following apparently correct equality: $$\mathop{\mathbb{E}}(X|Y<c)\cdot P(Y<c) = \mathop{\mathbb{E}}(\mathbb1_{(Y<c)}\cdot X)$$ where $c$ is a constant, and $P(Y<c)>0, P(Y\geq c)>0, \forall c$.
I believe this is true because of the law of total expectation: $$\mathop{\mathbb{E}}(X) = \mathop{\mathbb{E}}(X|Y<c)\cdot P(Y<c) + \mathop{\mathbb{E}}(X|Y\geq c)\cdot P(Y\geq c) $$ and the trivial equality: $$\mathop{\mathbb{E}}(X) = \mathop{\mathbb{E}}(\mathbb1_{(Y<c)}\cdot X) + \mathop{\mathbb{E}}(\mathbb1_{(Y\geq c)}\cdot X)$$
$\newcommand{\E}{\mathbb E}\newcommand{\d}{\mathrm d}\newcommand{\P} {\mathbb P}$ Edited Version. As user @Did noted, the first answer was misleading since OP's definition of conditional expectation given event was very unusual. One can be very short answering the question. The thing that needs to be proven follows directly from the definition since the expectation of $X$ given an even $A$ such that $\P(A) >0$ is: $$\E[X\mid A] =\frac{\E[X\mathbf{1}_A]} {\P(A)} $$ Multiplying both sides with $\P(A) $ yields the claim. We did not do anything.
First Version. WARNING: a very unusual definition is used here. Since I have respect for OP's definition I have decided to leave this answer undeleted. Both definitions seem to be equivalent after all.
Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space. So we have $X:(\Omega,\mathcal F)\to(E, \mathcal G)$ where $(E,\mathcal G)$ is a measure space. Let $A$ be an event such that $\mathbb P(A)>0$, then $\mathbb P(\cdot\mid A)$ is a well defined measure. You seem to define: \begin{align} \E[X|A]=\int_\Omega X(\omega)\, \d\mathbb P (\omega|A) \end{align} As said in the beginning: it is a rare definition, but here you go. One can check that $$\d\mathbb P(\omega|A)=\frac{\mathbf{1}_A}{\mathbb P(A)}\d\mathbb P(\omega)$$ We verify it with simple functions. Let $f(\omega)=\sum_{k=1}^N b_i \mathbf{1}_{B_i}(\omega)$: \begin{align} \int_\Omega f(\omega)\d\mathbb P(\omega|A)=\sum_{k=1}^N b_i \mathbb P(B_i|A)=\sum_{k=1}^N b_i \frac{\mathbb P(B_i\cap A)}{\mathbb P(A)}=\int_\Omega \sum_{k=1}^N b_i\frac{\mathbf{1}_{B_i\cap A }(\omega)}{\mathbb P(A)}\,\d\mathbb P(\omega)=\int_\Omega f(\omega)\frac{\mathbf{1}_A}{\mathbb P(A)}\d\mathbb P(\omega) \end{align} Since it holds for simple functions it holds for all functions (why?). Now we have: \begin{align} \E[X|A]=\int_\Omega X(\omega)\frac{\mathbf{1}_A}{\mathbb P(A)}\d\mathbb P(\omega)=\frac{\E[X\mathbf{1}_A]}{\mathbb P(A)} \end{align} A little bit more general result of your statement is proved.