Is there a commutative operation for which the inverse of the operation is also commutative? For instance addition is commutative, but the inverse, subtraction, is not.
$$
5+2 = 2+5\\
5-2 \neq 2-5
$$
Same for multiplication/division:
$$
5\times4 = 4\times5\\
5/4 \neq 4/5
$$
So is there a group operation $\circ$ with the inverse $\circ^{-1}$ such that
$$
a\circ b = b\circ a\\
a\circ^{-1}b = b\circ^{-1}a
$$

Part of the problem here is the difference between a function and an operation.
A function f(x) is defined as the mapping from objects x to their counterparts f(x).
The inverse function f-1(x) is the mapping that takes the objects back to where they started.
So f(f-1(x))=x
Under this definition, “adding” is not a function, but “adding two” is a function.
I believe that what you want is what I would describe as a binary operation. It is effectively a function of two values.
So the adding function can be written as add(x, y).
In this description commutativity means that add(x, y) = add(y, x)
I'm not sure that a binary operation can have an inverse in the proper sense, because how can we undo the addition to get the original pair of values? Given that add(x, y) = 6, for example, we can never determine which values x and y were added together.
So let's try to rephrase the question. I started thinking about addition and its “inverse” (or perhaps better “reverse”) subtraction.
Subtraction can also be written as subtract(x, y) and it is not commutative; subtract(x, y) is not equal to subtract(y, x).
The effect of subtraction is really to reverse the addition process.
So subtract(add(x,y), y) = x
The question is therefore if there are a pair of functions f and g that act on two values that have this property:
f(g(x, y), y) = f(g(x, y), x)
But if f is the reverse of g, that would mean that f(g(a, b), b) = a
So we would end up with y = x
Therefore the answer to your question is, “No.”
