Sure, supermodules over a superalgebra form a nice and nontrivial example of a tensor category. In fact, it is a theorem that Deligne that all tensor categories over $\mathbb{C}$ with suitable growth conditions can be obtained as categories of super-representations.
Another reason is given by Qiaochu, that supercommutativity arises naturally in a lot of places.
For instance, the exterior algebra is supercommutative; thus the wedge product on differential forms acts the same way. Interestingly, the exterior algebra of a direct sum corresponds to taking the super tensor product.
Another example: With singular cohomology on a topological space, there is a way to define a cup product, which satisfy the supercommutative law $ab = (-1)^{\deg b \deg a} ba$.
(Note that given a $\mathbb{Z}$-graded super-commutative algebra, you can get a $\mathbb{Z}/2$-graded superalgebra as you indicate in your question by taking the sum of the odd parts and the sum of the even parts.)