Let me just try a few more comments on this and on the answer of @Qiaochu Yuan:

One reasons for confusion might be the actual definition of the Picard group. If I understand you correct then you take iso classes of invertible bimodules without the assumption that the underlying algebra $A$ is commutative. In this case, one has a canonical short exact sequence
\begin{equation}
1 \longrightarrow \mathrm{InnAut}(A)
\longrightarrow \mathrm{Aut}(A)
\longrightarrow \mathrm{Pic}(A)
\end{equation}
and hence the outer automorphisms always contribute to the Picard group. IN particular, if this group is non-abelian, the Picard group is non-abelian as well. This is completely classical and can be found in all kind of books on Morita theory.

Now the catch comes when $A$ is actually commutative. In this case we can consider invertible bimodules which are in addition central: for $a \in A$ and $x \in E$ we have $ax = xa$, i.e. the left and right module structures coincide. If you think of a geometric situation then this is what you usually have in mind when taking about sections of line bundles: functions are multiplied to section from left or right without much thinking. But in principle you can twist the left module structure by means of an automorphism while keeping the right module structure.

Now taking only central invertible bimodules gives a subgroup of the Picard group, sometimes called the central/commutative/static Picard group $\mathrm{SPic}(A)$. This is the one which really corresponds to line bundles. It is always commutative! Moreover, in this case ($A$ commutative) one has a semi-direct product structure
\begin{equation}
\mathrm{Pic}(A) = \mathrm{Aut}(A) \ltimes \mathrm{SPic}(A)
\end{equation}
where the autormorphisms (are always outer!) act on the bimodules in the obvious way. So the full Picard group has this decomposition and hence its non-abelianness comes only from the automorphism group.

To make the confusion complete: in geometry people call $\mathrm{SPic}(A)$ sometimes the Picard group (since it corresponds to line bundles), which therefore is always abelian. This is perhaps the reasons that some people say that the Picard group is abelian.

A last remark: there are notions of Morita equivalence for $^*$-algebras as well, even beyond the $C^*$-algebraic case. Here you have similar statements on Picard groups (which now take into account $^*$-involutions and notions of positivity). You can find such things in a lecture note by myself on my homepage as well as in several papers also there.