Let $\mathcal{D}$ be a triangulated category, with objects $A, B \in \mathcal{D}$. Is it true that $$A \xrightarrow{0} B \to A[1] \oplus B \to A[1] \tag{$*$}$$ is a distinguished triangle?
If $\mathcal{D} = \mathcal{D}(\mathcal{A})$ is the derived category of an abelian category $\mathcal{A}$, one can just compute the mapping cone of the $0$-map, which turns out to be $A[1] \oplus B$.
The triangle $(*)$ looks like it is the direct sum of the two triangles $$\begin{array}{c} A & \to &0 & \to & A[1] & \xrightarrow{\operatorname{id}_{A[1]}} & A[1]\\ 0 & \to &B & \xrightarrow{\operatorname{id}_B} & B & \to & 0, \end{array}$$ but none of the axioms of a triangulated category state that direct sums of triangles are triangles. Or am I missing something here?
More generally, I wonder: Why does the definition of a triangulated category involve an additive category, but there is no connection between triangles and biproducts?
In a pre-triangulated category the direct sum of triangles is a triangle. A proof of this (and in fact a stronger statement) can be found at Lemma 13.4.9 of the Stack project.
Your question was also answered on Mathoverflow.