The answer to your question is no, in general. A simple counterexample is provided by a path algebra of the Dynkin quiver $D_4$ and another algebra tilted from it. That is, let $A$ be the path algebra where the quiver $D_4$ is oriented so that the vertex of degree 3 is a source and let $B$ be the quotient of the path algebra of the Euclidean quiver $\tilde{A}_3$--oriented so that we have two paths of length 2 from a source to a sink--by the relation expressing that these two paths are equal. Then $B$ is biserial (I believe it is even special biserial, if I remember the definition correctly), but $A$ is not biserial. That these algebras are derived equivalent follows from the fact that one is tilted from the other, which is remarked, for instance, in IV.4.3(b) of Happel's book *Triangulated categories in the representation theory of finite dimensional algebras*.

However, assuming that you mean to include Nakayama algebras (i.e., uniserial algebras) as (special) biserial, it is true for self-injective special biserial algebras: Pogorzaly shows that this class of algebras is closed under stable equivalence [Comm. Algebra 22 (1994), no. 4, 1127–1160], and derived equivalent self-injective algebras are always stably equivalent by a theorem of Rickard.