I am writing up as one answer the comments by @Johan, by @Libli, and by myself. If either of them prefers to write an answer, I am happy to delete this answer.
Let $A$ be an Abelian variety. For every scheme $S$ and every $S$-valued point $x\in A(S)$, denote by $\mu_x$ the associated translation automorphism of the $S$-scheme $S\times A$, i.e., $\mu_x(y) = x+y$.
Denote by $\widehat{A}$ together with the invertible sheaf $\mathcal{P}$ on $\widehat{A}\times A$ the relative $\text{Pic}^0$ of $A$, normalized so that $\mathcal{P}|_{\widehat{A}\times\{0\}}$ is the structure sheaf on $\widehat{A}$. Of course $\widehat{A}\times A$ is a commutative group scheme with its structure as the product of two commutative group schemes. Denote by $G$ the noncommutative group scheme structure on $\widehat{A}\times A$ defined by $$ ([\mathcal{L}],x)\bullet([\mathcal{M}],y) = ([\mathcal{L}\otimes \mu_x^*\mathcal{M}],x+y).$$ There is an "action" of $G$ on the bounded derived category of coherent sheaves on $A$ that associates to each $([\mathcal{L}],x)$ in $G$ and each bounded complex $C^\bullet$ of coherent sheaves on $A$ the associated bounded complex of coherent sheaves, $\mathcal{L}\otimes \mu_x^*(C^\bullet).$ (According to an article of Orlov, this action identifies $G$ with the identity component of the group of autoequivalences of the bounded derived category of coherent sheaves on $A$.)
If $C^\bullet$ is not quasi-isomorphic to the zero complex, i.e., if it is not an exact complex, there there exists an integer $p$ such that the cohomology sheaf $h^p(C^\bullet)$ is nonzero. In general, a "flat deformation" of the complex $C^\bullet$ does not necessarily give rise to a flat deformation of the coherent sheaf $h^p(C^\bullet)$, since base change is not left exact. However, for a connected, smooth group scheme $G$, for a $G$-equivariant family of deformations over $G$, the coherent sheaves $h^p$ are compatible with base change: this holds over a dense open of $G$ (since $G$ is reduced), and this dense open is $G$-invariant, thus it is all of $G$. Therefore, the deformations of $C^\bullet$ arising from the action of $G$ give rise to a deformation of $h^p(C^\bullet)$.
By hypothesis, the coherent sheaf $h^p(C^\bullet)$ on $A$ is nonzero. If the rank is positive, then the action of the normal subgroup $\widehat{A}\times\{0\}$ of $G$ on this sheaf is nontrivial by considering "det" of the coherent sheaf. If the rank of the sheaf is zero, i.e., if the support of the sheaf is a proper closed subscheme of $A$, then the action of the subgroup $\{[\mathcal{O}_A]\}\times A$ of $G$ on the sheaf is nontrivial since it "moves" this proper closed subscheme. Either way, the action of the group scheme $G$ on the sheaf is nontrivial.
Since the action of $G$ already produces nontrivial deformations of the sheaf $h^p(C^\bullet)$, it also produces nontrivial deformations of the complex $C^\bullet$. Thus, the only rigid complexes in the bounded derived category of $A$ are quasi-isomorphic to zero.