From Anton Fetisov's work I think we can extract a simple example. In fact this example should be universal for the class of examples he constructs.

Consider the free symmetric monoidal category on one object with duals. We can represent the objects of this category as a pair $(A^+, A^-)$ of finite sets, with a morphism $(A^+, A^-) \to (B^+, B^-)$ being a bijection between $A^+ \circ B^-$ and $A^- \circ B^+$, plus a finite number of loops. Morphisms are composed by gluing the diagrams and following the paths to form a new bijection, while also possibly creating loops.

Suppose we want to add (countably) infinite products by generalizing this to infinite sets. Then we run into trouble, because when we compose the paths, we could get something that goes around infinitely in a spiral rather than ending up at any exit point. But if we just define a morphism $(A^+, A^-) \to (B^+, B^-)$ to be a bijection between a subset of $A^+ \circ B^-$ and a subset of $A^- \circ B^+$, plus a set of loops, then we can compose morphisms as before, throwing out any infinite paths.

This satisfies the axioms of a symmetric monoidal abelian category, with union as the tensor product structure. The element $(\mathbb N,0)$ is idempotent and dual to $(0,\mathbb N)$, but not self-dual as $(\mathbb N,0)$ and $(0, \mathbb N)$ are not isomorphic.

From Anton Fetisov's work I think we can extract a simple example. In fact this example should be universal for the class of examples he constructs.

Recall that a connected second-countable one-manifold with boundary is a closed interval, an open interval, a half-open interval, or a circle.

Consider the category whose objects are pairs of countable sets $(A^+,A^-)$ and where a morphism $(A^+, A^-) \to (B^+, B^-)$ is a second-countable 1-manifold with boundary, whose boundary equals $A^+ \cup A^- \cup B^+ \cup B^-$, and such that each closed interval component has one boundary point in $A^+ \cup B^-$ and one in $A^- \cup B^+$ (up to isomorphism).

Composition of a morphism $(A^+ , A^-) \to (B^+, B^-)$ with a morphism $(B^+,B^-) \to (C^+,C^-)$ is obtained by gluing the manifolds along $B$.

A symmetric monoidal structure is given by disjoint union of sets and morphisms. The identity is a union of closed intervals connecting each element to itself.

If I did this with finite sets and compact manifolds, so that the only components were closed intervals and circles, I would have constructed the free symmetric monoidal category on one generator and its dual. Because I allow infinite sets, I must necessarily allow connected components that are not closed. 

The element $(\mathbb N,\emptyset)$ is idempotent as $(\mathbb N,\emptyset) \otimes (\mathbb N,\emptyset)= (\mathbb N \cup \mathbb N,\emptyset \cup \emptyset) = (\mathbb N , \emptyset)$.

It has dual $(\emptyset, \mathbb N)$, with the unit and counit each given by a union of closed intervals connecting the two copies of $\mathbb N$, where the relations hold by an infinite disjoint union of the usual stringy proof.

However, it is not self dual, as $(\mathbb N,\emptyset)$ and $(\emptyset,\mathbb N)$ are not isomorphic, as every map between them has all elements connected to half-open intervals (because $B^- \cup A^+$ is empty), so the composition also has all elements connected to half-open intervals and thus isn't the identity.