Here is half of a classification. Let $\otimes$ be a monoidal structure on $\Delta_+$. As I mentioned in a comment, the monoidal unit must be $[-1]$ or $[0]$ because these are the only objects with commutative endomorphism monoids.

Suppose that the monoidal unit is $[0]$. Let us consider $[1] \otimes [1]$. We have that $[0]$ is a retract of $[1]$ in 2 ways, and as a result we obtain 4 retracts of $[1] \otimes [1]$ with support $[0] \otimes [0] = [0]$. Consider the induced linear ordering on these 4 points. We also have 4 ways that $[1] = [1] \otimes [0] = [0] \otimes [1]$ is a retract of $[1] \otimes [1]$, and from this we can deduce most of the ordering. It must have the following relations

$\require{AMScd} \begin{CD} 0 \otimes 0 @>>> 0 \otimes 1\\ @VVV @VVV\\ 1 \otimes 0 @>>> 1 \otimes 1 \end{CD}$

To complete this to a linear order, without loss of generality we must have $0 \otimes 1 \leq 1 \otimes 0$. But now, one of our 4 projections onto $[1]$ is the coordinate projection onto the right column of the above square. The fact that this projection is order-preserving implies that $1 \otimes 0 = 0 \otimes 1 = 1 \otimes 1$. This contradicts the fact that the right column exhibits $[1]$ as a retract of this subset of $[1] \otimes [1]$.

Therefore the monoidal unit is not $[0]$; it must be $[-1]$.

I also think I'm ready to conjecture that $\oplus$, $\oplus^{rev}$ (as mentioned by Peter) and the degenerate monoidal structure are probably the only ones. You could imagine a classification starting as follows. Consider the maps $[0] = [0] \otimes [-1] \to [0] \otimes [0]$ and $[0] = [-1] \otimes [0] \to [0] \otimes [0]$. If these are the same, then we should have the degenerate monoidal structure. If they are different, then one is less than the other, and those two cases should correspond to $\oplus$ and $\oplus^{rev}$.

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