Here is a crude way to see that this is true: Given $a\in A$ and $b\in B$, consider $A-a =\{x-a | x\in A\}$ and $b-B =\{b-y | y\in B\}$. These shifted triangles have a convex intersection that is more than a point because $2+2 \gt 3$. The intersection gives the pairs of points adding up to $a+b$, and each extreme point of the intersection corresponds to an extreme point of $A$ or $B$, hence a point contained in an edge of $A$ or an edge of $B$.

I think there is a nicer argument in terms of the projection of $A \times B$ to the Minkowski sum.

Here is a crude way to see that this is true: Given $a$ and $b$ in the interiors of $A$ and $B$, respectively, consider $A-a =\{x-a | x\in A\}$ and $b-B =\{b-y | y\in B\}$. These shifted triangles have a convex intersection that is more than a point because $2+2 \gt 3$ so the tangent spaces have nontrivial intersection. The intersection $\{x-a=b-y\}$ gives the pairs of points $\{(x,y) | x+y=a+b\}$ adding up to $a+b$. Each extreme point of the intersection corresponds to an extreme point of $A$ or $B$, hence a point contained in an edge of $A$ or an edge of $B$.

This argument only used that $A$ and $B$ are convex, so the same argument works for other convex $2$-dimensional shapes in $\mathbb{R}^3$, or $d$-dimensional shapes in $\mathbb{R}^{2d-1}.$

There is a nicer version of this argument in terms of the map from $A \times B$ to the Minkowski sum. The preimage of a point must intersect the $3$-skeleton which consists of $(A\times \delta B) \cup (\delta A \times B).$