It seems it will be possible only when all four angle bisectors of the quadrilateral are coincident (all intersecting in one point). The reason for this is that given any angle in the quadrilateral, its bisector contains all and only those points equidistant from the lines containing the sides of the angle. So given that there are four angles, any point shared by their angle bisectors will be equidistant from each of the sides of all respective angles, and any point not on an angle's bisector will not be equidistant from its sides.
Figure 1 shows a quadrilateral for which there is not one point equidistant from its four sides. One point (G) is equidistant from DA, AB, and BC, but this distance is not the same as the distance from G to DC. Another point (E) is equidistant from AD, DC, and BC, but this distance is not the same as the distance from E to AB. The problem is that G and E don't coincide.
Figure 1.
The question now is, "For what quadrilaterals are angle bisectors coincident?" Not all of them, true. But which of them?
It seems clear that angle bisectors will be coincident for squares and rhombi. An argument for this would probably aim to conclude that bisectors of opposite angles would pass through the opposite vertices, so they would necessarily coincide.
I think a kite also would have the property that two of its angle bisectors would coincide, and the other two bisectors would intersect on the coincident bisectors (see Figure 2). The argument in support of this would probably use congruent triangles or something. I'm not going to argue that now because there's one more case that really surprised me.
Figure 2.
The really surprising thing is that that there are scalene quadrilaterals for which there exists a point equidistant from all four sides (see Figure 3 for an example). This isn't to say that each scalene quadrilateral has a point equidistant from every side. Rather, there are lots of scalene quadrilaterals for which this is true.
Figure 3.
The key idea seems to be that if you have three non-collinear points A, B, and C, you can locate a fourth point (D) so that angle bisectors coincide. I haven't yet figured out a construction that will actually locate such a point, but it seems clear to me that there are an infinite number of them.
For example, in Figures 4a through 4d I've shown how you can start with points A, B, and C, and then move point D so that eventually the angle bisectors coincide. What I don't know, yet, is how to construct point D.
Figure 4a
Figure 4b
Figure 4c
Figure 4d
At first I thought the problem would reduce to "Given two points which will be on the sides of an angle and a point which will be on the that angle's bisector, construct such an angle." If I could construct such an angle, then I could complete a quadrilateral by starting with one angle and a point in its interior on the angle's bisector. My problem was that I couldn't locate a vertex. I tried examining all kinds of ratios to see what must be true of any vertex of such an angle, hoping that I could then work backward from what must be true to what I needed.
I then thought, "Suppose I start with more of the quadrilateral given, such as three adjacent vertices and a point on the bisector of the quadrilateral's angle formed by those three vertices" (Figure 5).
Figure 5.
Starting with these elements then made the construction easy! Since Pt on side1 will be a vertex, and the bisector of the angle having it as vertex must pass through Pt on Bisector, we can find the angle for which Pt on side1 is the vertex by reflecting Angle vertex through the line containing Pt on side1 and Pt on Bisector. This then gives a point on the other side of the angle for which Pt on side1 is vertex. Do the same for Pt on side2, getting a point on the other side of the angle for which Pt on side2 is the vertex. The intersection of those two sides gives the fourth vertex of a quadrilateral which has the property that all four angle bisectors pass through Pt on Bisector (see Figure 6).
Figure 6
Finally, if we examine the locus of Fourth Point, it appears any fourth point lies on a hyperbola having some relation with the original 3 vertices (see Figure 7). At this time I don't see the precise relationship.
What we have, now, is a solution to a problem similar to the original one, but different in several important ways. The original question asked for a method to construct a point equidistant from the sides of a given quadrilateral. There is no solution to the general case, since there will not be a point equidistant from all four sides for any quadrilateral not having coincident angle bisectors.
However, we did construct the class of quadrilaterals for which there is a point equidistant from all four sides under the condition we are given three vertices. We noticed that the fourth vertex can lie on what seems to be a hyperbola. The definition of that hyperbola still remains to be given independently of constructing the fourth vertex.