This is a most interesting question. I pondered on it for awhile, and here's what I could come up with :
Here are the basic stipulations. One is given two unequal line segments, length a and b (without loss of generality, a > b). With this alone, and with the aid of unmarked ruler and compasses, what line segments can be constructed ?
Immediately, one can see how to work out (a + b) and (a - b) : just superimpose the lines and extend or demarcate as needed. In fact, any linear combination of length (ma + nb) can be worked out for integral m and n.
Now, one can also bisect ad nauseam to give all lengths of the form \frac{ma}{2^p} + \frac{nb}{2^q} where m and n are integers and p and q are naturals.
Further, using http://www.jimloy.com/geometry/trisect0.htm ingenious method of trisecting a line segment, we can extend the list of possibilities to : \frac{ma}{(2^p)(3^r)} + \frac{nb}{(2^q)(3^s)} I don't know if extension to further prime divisors is possible, if I'm not mistaken, Gauss came up with some results in this area.
Now, let's go to roots. Interestingly, \sqrt{ab} can be fairly easily constructed, like so :
(a+b)^2 - (a-b)^2 = 4ab so what we need to do is to construct line segment of length (a-b), then use one end to draw a circle of radius (a+b). From the other end, we need to construct a perpendicular to this line segment, and connect the intersection of the perpendicular and the circle to the first point. This is the hypotenuse of a right triangle and it has length 2\sqrt{ab}. Bisect that to get \sqrt{ab}. So now, we can get both the arithmetic mean and the geometric mean of two line segments.
From this we can get various combos of the form : \sqrt{ma^{p/2}b^{q/2} + na^{r/2}b^{s/2} and work up to higher binary exponents. The problem here is that neither the addition nor either of the a or b terms can be eliminated. Only composites involving sums of products of various binary roots of a and b can be gotten.
If we are given an additional piece of critical info, namely, the length of the unit line segment (in the form of a line segment which is defined to be of unit length), then we can do a lot more very easily.
Given a, b and 1 (unity), immediately, a/b and b/a can be seen as gradients on a Cartesian plane where the a and b segments are marked out perpendicular to each other. To get line segments corresponding to those just mark out unity on either the vertical or horizontal axis, produce and voila !
Reciprocals (1/a, 1/b, 1/ab, and other combos) are also easy. The harmonic mean of a and b can now also be constructed.
Similarly, products like (ab), and all integral powers of a and b individually and combos of the form a^mb^n with m and n integral (negatives too) can be gotten. Using the previous results, combos of binary roots and so forth can also be derived.
