As the ratio r express the absolute value of ratio of the segments AP/BP, there are two point M and N on the straight line AB which satisfy the above condition. Indeed r and minus r are the barycentric coordinates of M and N with respect to to A and B on line AB or in other words the simple ratio ABM and ABN are equal an opposed in sign.
Therefore the cross ratio < ABMN> is harmonic being M and N conjugated.

So if we draw a pencil of rays from any arbitrary point in the plane to the points A,B,M, and N the harmonic ratio will be preserved, however since point P must be selected so ray PM be the bisector of angle APB, the ray PN must be the other exterior bisector of the angle APB. All the above means that the angle NPM must be always 90 degrees and since MN is constant P must belong to a circumference of diameter MN.

I do not see a flaw, no.

I can't say whether it is simpler or more elementary. What you do is replace the ratio property of angle bisectors (two of them being used simultaneously) with a property of harmonic ratio which also tackles two ratios simultaneously. Although it all depends on a view point, I would say that your suggestions is as simple as the proof on thepage but less elementary.

For what it is worth...

However, bringing in the harmonic ratio may suggest a generalization to the locus of a fixed cross-ratio. So it is always beneficial to have a second view.

>As the ratio r express the absolute value of ratio of the
>segments AP/BP, there are two point M and N on the straight
>line AB which satisfy the above condition. Indeed r and
>minus r are the barycentric coordinates of M and N with
>respect to to A and B on line AB or in other words the
>simple ratio ABM and ABN are equal an opposed in sign.
>Therefore the cross ratio < ABMN> is harmonic being M and N
>conjugated.
>
>So if we draw a pencil of rays from any arbitrary point in
>the plane to the points A,B,M, and N the harmonic ratio will
>be preserved, however since point P must be selected so ray
>PM be the bisector of angle APB, the ray PN must be the
>other exterior bisector of the angle APB. All the above
>means that the angle NPM must be always 90 degrees and since
>MN is constant P must belong to a circumference of diameter
>MN.