ag.algebraic geometry - theta divisor on a principally polarized abelian variety

This is true. For $A_{/mathbb{C}}$ an abelian variety, $L$ an ample line bundle on $A$, then any line bundle
$M in operatorname{Pic}^0(A)$ -- over $mathbb{C}$, this is equivalent to having first Chern class zero -- is of the form $T_x^{*} L otimes L^{-1}$ for some $x in A$. (e.g. Theorem 1 on p. 77 of Mumford's Abelian Varieties).

Applying this theorem with $L = L(D_2)$, $M = L(D_1) - L(D_2)$, we get that

$L_1 - L_2 = T_x^*(L_2) - L_2$, so

$L_1 = T_x^*(L_2)$.

So $D_1$ and $x+D_2$ (meaning translation of $D_2$ by $x$!) must be linearly equivalent, but by your assumption $h^0(L(D_1)) = h^0(L(D_2)) = 1$, they are each the unique effective
divisors in their linear equivalence classes, so we must have $D_1 = x + D_2$.