= Solution
Let $m_I:X^3\to X$ add the coordinates indexed by a nonempty subset $I$ of the three-element index set. The <Theorem of the Cube> says that for every line bundle $\mathcal L$ on an abelian variety,
$$
m_{123}^*\mathcal L
\otimes m_{12}^*\mathcal L^\vee
\otimes m_{13}^*\mathcal L^\vee
\otimes m_{23}^*\mathcal L^\vee
\otimes m_1^*\mathcal L
\otimes m_2^*\mathcal L
\otimes m_3^*\mathcal L
$$
is trivial, up to the harmless constant line given by the fiber of $\mathcal L$ at the identity.
Pull this line bundle back along $(f,g,h):Y\to X^3$. Pullback commutes with tensor products and duals, and $m_I\circ(f,g,h)$ is the corresponding sum of morphisms. The resulting bundle is precisely $\mathcal M_{f,g,h}$, so it is trivial.
Take $Y=X$, $f=\operatorname{id}_X$, and let $g,h$ be the constant maps with values $x,y$. All pullbacks along constant maps are trivial line bundles. The formula for $\mathcal M_{f,g,h}$ then becomes
$$
T_{x+y}^*\mathcal L\otimes
(T_x^*\mathcal L)^\vee\otimes
(T_y^*\mathcal L)^\vee\otimes\mathcal L
\simeq\mathcal O_X,
$$
or equivalently
$$
T_{x+y}^*\mathcal L
\simeq T_x^*\mathcal L\otimes T_y^*\mathcal L\otimes\mathcal L^\vee.
$$
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