Number Partitioning targeting a Ratio of subset sums?
Yes.  Suppose the numbers are $x_1,\dots,x_n$ and the target ratio is $r$.  Define

$$y=\left\lfloor {1-r \over 1+r} \times (x_1 + \dots + x_n) \right\rceil,$$

and then look for the partition of $x_1,\dots,x_n,y$ that has minimum difference.  I suspect this might be either the solution to your original problem, or very close to it.

Why?  A partition of $x_1,\dots,x_n$ with exactly the ratio $r$ will have one group whose total is $c (x_1+ \dots + x_n)$ and another whose total is $cr(x_1+\dots+x_n)$.  These two must sum to $x_1+\dots+x_n$, so we find that $c+cr=1$, i.e., $c=1/(1+r)$, so their difference is

$$c(x_1+\dots+x_n) - cr(x_1+\dots+x_n) = {1-r \over 1+r} (x_1+\dots+x_n).$$

We have defined $y$ to be this difference, rounded to the nearest integer.  So, if you add $y$ to the smaller group, you get a perfect partition of $x_1,\dots,x_n,y$ with difference zero.  And the converse is true as well.