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.