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 inclusion homotopy pushout of $S A \setminus S U \to S A$ is a homotopy equivalence, while and $S X \setminus S U U$ along $S A \to setminus S X$ isn't, so U$has the square in question is not a homotopy pushouttype of the wedge of three circles, so it cannot be$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$in the corresponding triple. 1 The answer to your first question is negative. Before I give a counterexample, let me rephrase the problem in terms I consider more natural. First, I believe it is more convenient to consider excisive triads instead of excisive triples, i.e. I will replace a triple$(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 inclusion$S A \setminus S U \to S A$is a homotopy equivalence, while$S X \setminus S U \to S X$isn't, so the square in question is not a homotopy pushout. 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.