We show that these two spaces are both homeomorphic to $I \times X \times Y / \left(I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)\right)$ as follows:
- Passing from $I \times X \times Y$ to $X * Y$, we identify the points (0,x,y) \sim(0,x,y') and (1,x,y) \sim (1,x',y). Under these identifications, $I \times X \times \{y_0\} \cup I \times \{x_0\} \times Y$ becomes $X *\{y_0\} \cup \{x_0\}*Y$. Hence $\frac{I \times X \times Y } {I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)} \simeq \frac{X * Y}{X * \{y_0\} \cup \{x_0\}*Y}$
- Notice that $ \frac{I \times X \times Y } {I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)} \simeq \frac{I \times X \times Y } {I \times ( X \vee Y)} \simeq I \times \frac{ X \times Y } { X \vee Y} \\ \simeq I \times(X \wedge Y) \simeq I \times \frac{ X \wedge Y } {\{x_0\}\wedge \{y_0\}} \simeq \frac{I\times ( X \wedge Y) } {I \times (\{x_0\}\wedge \{y_0\})} .$ When passing form $I\times ( X \wedge Y) $ to $S(X \wedge Y)$, we collapse $\{0\} \times (X \wedge Y)$ and $\{1\} \times (X \wedge Y)$ to two points. With these identifications, $I \times (\{x_0\}\wedge \{y_0\})$ becomes $S(\{x_0\}\wedge \{y_0\})$. Whence $\frac{I \times X \times Y } {I \times ( X \times \{y_0\} \cup \{x_0\} \times Y)} \simeq \frac{S(X\wedge Y)}{S(\{x_0\}\wedge \{y_0\})}.$
Since $({X * Y},{X * \{y_0\} \cup \{x_0\}*Y})$ and $({S(X\wedge Y)},{S(\{x_0\}\wedge \{y_0\})})$ are CW-pairs, they both have homotopy extension property (Hatcher prop. 0.16), so the complex is homotopy equivalent to the quotient space by the subcomplex (Hatcher prop. 0.17). Therefore ${X * Y}\simeq {S(X\wedge Y)}.$