FPRAS for Perfect Matching
No. Knowing $x$ and $y$ to within multiplicative error bounds of $1\pm\varepsilon$ does not give you a multiplicative error bound on $x-y$. In particular, if $x=y$, being allowed only multiplicative error means that an FPRAS for $x-y$ would have to return exactly zero with probability at least $\tfrac34$, whereas you could end up returning any value $v$ such that $|v|\leq 2\varepsilon\cdot |x|$.