I could only prove one direction, let $A \in \sigma(Y)$
$$\int_A\mathbb{E}[f(X)\mid Y] d\mathbb{P} =\int_Af(X) d\mathbb{P} = \mathbb{E}[f(X) \mathbb{1}_A] = \mathbb{E}[f(X)] \mathbb{E}[ \mathbb{1}_A] = \int_A \mathbb{E}[f(X)]d\mathbb{P}$$
for third equality I used independence and every constant is measurable. so...
how to do the reverse implication though ?
For the opposite implication, let $A$ and $B$ be two Borel subsets of the real line. Using the assumption where $f$ is the indicator function of $A$, we get $$\tag{*} \mathbb E\left[\mathbf 1_{A}\left(X\right)\mid Y\right]=\mathbb P\left(X\in A\right). $$ By using the definition of conditional expectation with $\{Y\in B\}\in\sigma(B)$, we get that $$ \mathbb E\left[\mathbb E\left[\mathbf 1_{A}\left(X\right)\mid Y\right]\mathbf 1_B\left(Y\right)\right]=\mathbb E\left[ \mathbf 1_{A}\left(X\right)\mathbf 1_B\left(Y\right)\right]$$ and replacing $\mathbb E\left[\mathbf 1_{A}\left(X\right)\mid Y\right]$ by $\mathbb P\left(X\in A\right)$ gives the independence between $X$ and $Y$.
Observe also that a similar criterion can be established in order to show the independence between $X$ and a $\sigma$-algebra $\mathcal G$, namely, that for each measurable bounded function $f$, the equality $$ \mathbb E\left[f(X)\mid\mathcal G\right]=\mathbb E\left[f(X) \right] $$ holds almost surely.