$f(P)=f(Q)$ implies that $P=Q$ Let $(X,\mathbb{H})$ and $(Y,\mathbb{F})$ be two measurable spaces. Assume that $P$ and $Q$ be probability measures on $(X,\mathbb{H})$ and that $f:X\to Y$ is a $\mathbb{H}/\mathbb{F}$-measurable mapping. What are the weakest (alternatively some weak) conditions on $f$ for which
$$
f(P)=f(Q) \implies P=Q,
$$
holds? Here $f(P)$ and $f(Q)$ are push-forward measures.
If we work under the assumption that $X$ and $Y$ are metric spaces and $P,Q$ are Borel measures then it is sufficient to say that $f$ is a homeomorphism, but what does this translate to when we have no topology on $X$ and $Y$ only $\sigma$-algebras.

What you are struggling with is the measure determination problem and I do not think that 
it is fruitful to try finding a function $f$ such that $f(P)=f(Q)\Rightarrow P=Q$. A family $\mathscr{M}$ of bounded $\mathbb{R}$-measurable functions is said to be measure determining whenever two finite measures $P$ and $Q$ on $(X,\mathbb{H})$ satisfy 
$$
\int \varphi dP=\int \varphi dQ\quad (\forall \varphi \in  \mathscr{M}), 
$$
then $P=Q$. Let me show some examples. 


*

*Let $\mathscr{P}$ be a $\pi$-class (i.e., a class that is closed under finite intersection) 
that generates $\mathbb{H}$. Then, 
$\{1_{A}:A\in \mathscr{P}\}$ is measure determining (Billingsley, Probability and Measure, 
3rd ed., p.163, Theorem 10.3 or Kallenberg, Foundations of Modern Probability, 2nd ed., p.9, 
Lemma 1.17), where $1_A$ is the indicator function of the set $A$. 
If $X$ is a topological space and $\mathbb{H}$ is generated by the topology, 
then trivially the family of open (or closed) sets is a $\pi$-class generating 
$\mathbb{H}$ (the result you found is reduced to this case). If $X$ is a second countable topological space, a countable base $\mathscr{B}$ is a $\pi$-class 
that generates the Borel-$\sigma$-algebra. 
If $X=\mathbb{R}$, then $\{(-\infty ,x]:x\in \mathbb{R}\} $ is a $\pi$-class that generates the Borel-$\sigma$-algebra. 

*If $X$ is a metric space, the family of bounded, uniformly continuous $\mathbb{R}$-valued functions is measure determining. This can be proved by using the the dominated convergence theorem. 
