Thanks for pointing this out. You're definitely right. I hadn't checked this carefully since it was only to do the counterexample in model-independent way, but I totally messed up the combinatorics! I think the right expression is that
X₁₊ × ⋯ Xₙ₊ ≃ ∨_{∅ ≠ I ⊂ {1,…,n}} (∏_{i ∈ I} Xᵢ)₊ .
So if you split off the I = {1,…,n} and #I = 1 terms in this expression you get
X₁₊ × ⋯ Xₙ₊ ≃ (∏_{i ∈ I} Xᵢ)₊ ∨ (∐_{i ∈ I} Xᵢ)₊ ∨ [∨_{I ⊂ {1,…,n}, 1 <#I <n} (∏_{i ∈ I} Xᵢ)₊ ].
(So the two terms I wrote, plus a bunch of other junk.)