If the central series contains only G and e, then G is abelian. Every normal subgroup intersects the center of G, which is all of G.
Proceed by induction on the length of the central series.
Let C be the center of G and let K be a disjoint normal subgroup. Let D be the direct product of C and K. Since C commutes with everything, the subgroup D, generated by C and K, is indeed the direct product of C and K.
Take x in C and y in K, and conjugate xy by z, for any z in G. This yields x times the conjugate of y. Since K is normal the conjugate of y lies in K, and the conjugate of xy lies in D, hence D is normal in G.
Mod out by C, and D maps to a normal subgroup of G/C. In fact D maps to a copy of K. More than that, K maps 1-1 onto a copy of K.
We know G/C has a nontrivial center, since G is nilpotent. By induction, some element z is in the center of G/C, and in the image of K, which is normal in G/C. Pull z back to an element y in G that lies in K. In other words, y represents a coset of C that is in D, and is also in the center of G/C.
Select any x in G, and xy/x appears somewhere in K. Map this forward, from G into G/C. Since y winds up in the center, xy/x becomes y in the image. Thus xy/x is in the same coset as y. Write this as xy/x = wy, where w is in the center of G. But remember, xy/x is in K, which is e cross K in the direct product. Therefore w = e, and xy = yx. This holds for all x, so y is in the center after all.
We have arrived at a contradiction. The subgroups C and K cannot be disjoint. Every normal subgroup intersects the center of a nilpotent group.