Sure: a slight enhancement of Geoff Robinson's comment will work. Let $s$ and $t$ be reflections (I'm going to write reflections instead of pseudoreflections) with the same image in the abelianization. Then for every one dimensional character $\chi$, we have $\chi(s)=\chi(t)$. Given a $G$-orbit of reflecting hyperplanes, there is a linear character $\chi$ defined as follows: take a linear form $\alpha_H$ defining each hyperplane $H$, and let $f$ be the product over the orbit of all these forms (in other words, a generator for the ideal of polynomial functions vanishing on the union of the hyperplanes in the orbit). Then $g(f)=\chi(g) f$ for all $g \in G$.

This character has the property that $\chi(g)=1$ if $g$ is a reflection in a hyperplane not belonging to the orbit, while $\chi(g)=\mathrm{det}(g)^{-1}$ for $g$ a reflection in a hyperplane in the orbit. So $s$ and $t$ must be reflections in hyperplanes in the same $G$-orbit. Supposing that $s$ is a reflection in a hyperplane $H$ and $t$ is a reflection in a hyperplane $H'$, let $s_{H'}$ and $s_H$ be generators for the pointwise fixers of $H$ and $H'$ in $G$---we may and will assume that $s_{H'}$ and $s_H$ are $G$-conjugates. Now $s=s_H^\ell$ and $t=s_{H'}^k$ for integers $\ell$ and $k$; these integers are determined by the determinants of $s$ and $t$, so by assumption $k=\ell$. Since $s_H$ and $s_{H'}$ are conjugate, this does it.

Here is a way to codify what we have done: write $\mathcal{A}$ for the set of reflecting hyperplanes of $G$ and for such a hyperplane write $G_H$ for its pointwise fixer. There is a map
$$\{\text{group homomorphisms} \ G \rightarrow \mathbb{C}^\times \} \rightarrow \{\text{group homomorphisms} \ \prod_{H \in \mathcal{A}} G_H \rightarrow \mathbb{C}^\times \}^G$$ given by restriction (I hope the notation on the target here is self-explanatory). The previous two paragraphs amount to showing that for any finite linear group $G$ this map is surjective; this doesn't really depend on $G$ being a reflection group. That it is also injective follows if the subgroups $G_H$ generate $G$, or in other words if $G$ is a reflection group.