The answer to your first question is negative. Before I give a counterexample, let me rephrase the problem in terms I consider more natural.

First, I believe it is more convenient to consider excisive triads instead of excisive triples, i.e. I will replace a triple $(X, A, U)$ by a triad $(X; A, B)$ where $B = X \setminus U$.

Second, the excision (either in homotopy or homology) is not really a statement about excisive triples or triads, but about homotopy pushouts. Excisive triad is just a model for homotopy pushout with some specific point-set properties, which make topological arguments possible. By this I mean that $X$ is a homotopy pushout of $A$ and $B$ along $A \cap B$. (I don't think it is literally true that every excisive triad is a homotopy pushout, but those that aren't should be considered pathological anyway. However, every homotopy pushout is homotopy equivalent to an excisive triad.)

Thus your question could be rephrased as follows: given an excisive triple $(X, A, U)$ is the triad $(S X; S A, S X \setminus S U)$ excisive or at least a homotopy pushout? As you observed this triad is not excisive, which doesn't really tell us much since it still could be a homotopy pushout. However, this also doesn't have to be true. Let $X = S^1$ (as a subspace of $\mathbb{C}$ to fix the notation), $U = \{-1, 1\}$ and $A = X \setminus \{-i, i\}$. You can write down the suspended triad and observe that the homotopy pushout of $S A$ and $S X \setminus S U$ along $S A \setminus S U$ has the homotopy type of the wedge of three circles, so it cannot be $S X$.

On the other hand, it is easy to see that given an excisive triad $(X; A, B)$, the triad $(S X; S A, S B)$ is again excisive, which seems like a more natural thing to expect.

To answer your second question, I don't know the book you mention, but I assume that the proof of the Homotopy Excision Theorem is more or less the same as in tom Dieck's *Algebraic Topology*. In this proof the only moment when the point-set properties of $A$ and $B$ are used is when we map a cube into $X$ and use the Lebesgue Lemma to subdivide it into cubes mapping into $A$ or $B$. To do this we only need to assume that interiors of $A$ and $B$ cover $X$. This is equivalent to saying that closure of $U$ is contained in the interior of $A$ in the corresponding triple.