If a symmetry pairs two vertices, then the angles at those vertices must be congruent. A symmetry that pairs adjacent vertices of a parallelogram requires that the figure be a rectangle: adjacent angles are supplementary; if they're also congruent, then they must be right angles.
Having ruled out rectangles, your proposed symmetry must pair each vertex (a) with its non-adjacent counterpart, or (b) with itself. In (a), the axis would be the perpendicular bisector of the diagonal between the paired vertices; in (b) the axis would contain the vertex. If both (a) and (b) occur, then we have a vertex (from (b)) on the perpendicular bisector of the diagonal joining two non-adjacent others (from (a)); this implies that we have two congruent adjacent edges, which in turn implies that the figure is a rhombus.
To avoid both rectangles and rhombi, we must have that the axis provides only-(a) symmetry, or only-(b) symmetry. An axis of only-(a) symmetry must be the perpendicular bisector of both diagonals, which (as perpendiculars of that axis, and lying in the same plane of that axis) must therefore lie on the same line: the vertices of the figure are collinear. An axis of only-(b) symmetry contains all the vertices, making them collinear (but in a different arrangement than required by (a)).
If you disallow "degenerate" parallelograms with all vertices collinear, then you have your impossibility argument. However, I prefer to recognize degenerate figures as legitimate, and I try to avoid stigmatizing them whenever possible. (After all, they come in handy as intermediate steps in transforming one figure smoothly into another. Besides, it's not like the "Parallelogram Law of Vector Addition" becomes invalid when the vectors involved are linearly dependent.) So, I would go so far as to observe that an axis of only-(a) symmetry implies the existence of a (separate) axis of only-(b) symmetry, and vice versa; that is, we've determined a third class of parallelograms --along with rectangles and rhombi-- that have two axes of symmetry.
(Consideration of parallelograms whose vertices coincide is left to the reader.)
Note: An only-(a) axial symmetry can also be realized with an axis perpendicular to the plane of the parallelogram (which need not be degenerate), but you've ruled this out.