If you are asking for a simple mathematical identity in addition, subtraction, multiplication and division, then I am afraid that I do not know of any other than those already mentioned. Since you mentioned, however, that you were asking "merely for curiosity's sake", I will attempt to satisfy at least some of that curiosity.
Since bitwise operations are, after all, iterative, there exists for all bitwise operations a sequential sum that, for each given input, returns an identical output to that of the bitwise operation. Unfortunately, these are computationally incredibly expensive, but will serve all the same as a purely mathematical solution.
The mathematical equivalents to right- and left-shifts are already widely known, so I will not repeat them here. The simplest of the sequential sums that I could find, following these, was that of a bitwise NOT. While you have already stated that it can be expressed as $-1 - n$, this is more of a hack exploiting the fact that $-1$ (in 32-bit) is equal to 0xFFFFFFFF (hexadecimal, continuing your precedent of c-style notation). NOT, mathematically, can be expressed as follows:
$$
\operatorname{not}(a)=\sum_{n=1}^{\left \lfloor \log_{2}(a) \right \rfloor} \left [ n \left ( 1 - \left ( \left \lfloor \frac{a}{2^{n}} \right \rfloor \bmod 2 \right ) \right ) \right ]
$$
I will now move fairly quickly through the remaining bitwise operations.
OR:
$$
\operatorname{or}(a,b)=\sum_{n=1}^{\left \lfloor \log_{2}(a) \right \rfloor} \left [ n \left \lceil \frac{ \left ( \left \lfloor \frac{a}{2^{n}} \right \rfloor \bmod 2 \right ) + \left ( \left \lfloor \frac{b}{2^{n}} \right \rfloor \bmod 2 \right ) }{2} \right \rceil \right ] + b - b \bmod 2^{\left \lfloor \log_{2}(a) \right \rfloor}
$$
AND (using the floor of an addition):
$$
\operatorname{and}_1(a,b)=\sum_{n=1}^{\left \lfloor \log_{2}(a) \right \rfloor} \left [ n \left \lfloor \frac{ \left ( \left \lfloor \frac{a}{2^{n}} \right \rfloor \bmod 2 \right ) + \left ( \left \lfloor \frac{b}{2^{n}} \right \rfloor \bmod 2 \right ) }{2} \right \rfloor \right ]
$$
AND (using multiplication):
$$
\operatorname{and}_2(a,b)=\sum_{n=1}^{\left \lfloor \log_{2}(a) \right \rfloor} \left [ n \left ( \left \lfloor \frac{a}{2^{n}} \right \rfloor \bmod 2 \right ) \left ( \left \lfloor \frac{b}{2^{n}} \right \rfloor \bmod 2 \right ) \right ]
$$
XOR:
$$
\operatorname{xor}(a,b)=\sum_{n=1}^{\left \lfloor \log_{2}(a) \right \rfloor} \left [ n \left ( \left ( \left ( \left \lfloor \frac{a}{2^{n}} \right \rfloor \bmod 2 \right ) + \left ( \left \lfloor \frac{b}{2^{n}} \right \rfloor \bmod 2 \right ) \right ) \bmod 2 \right ) \right ] + b - b \bmod 2^{\left \lfloor \log_{2}(a) \right \rfloor}
$$
I am aware of the fact that this thread has not seen any activity in a while, but I myself spent some time looking (unsuccessfully) on the internet for this before deciding to work it out myself. Since I did work this out by hand, however, there may be errors in the formulas. If anyone finds any, please let me know of them, and I will attempt to fix them.
Note that I could not find any mathematical way of performing these operations without the use of modulus, floors, ceilings and the sequential additions. If anyone finds any alternative methods of solving these, I would be interested in hearing of them (hypothetically, substitution of the equation for triangular numbers into the sequential sums may be able to simplify them).
Also note that, programatically, the floor and ceiling functions can be replaced by integer operations.
Hopefully somebody, sometime, will find a use for these equations.
In case anybody is wondering, the reason why I put these together was to discover if it is possible to apply calculus to bitwise operations. It may or may not be, depending on whether or not sequential sums, floors, ceilings and modulo operations can be used in calculus.
Mathematics and physics are not really compatible.
In the plane, or the 3D space (and so forth) it is fine to represent a vector space as a magnitude and direction.
However, the formal definition of a vector space requires not the need for either of those in order to represent a vector. In fact, a vector - formally - is just an element of a vector space.
This can go on, not all vector spaces have norms defined on them. Without the axiom of choice, not all have a basis, decomposition into a direct sum, nontrivial functionals, and so on.
Similarly, not all topological spaces are normal, regular, Hausdorff, etc., however we like to think of the physical world as $\mathbb R^3$ which is normal, regular, Hausdorff, etc..
In the finite dimensional case, or assuming the axiom of choice, we have a basis for the space. That is every vector can be written as a linear combination of the elements of the basis. You can think of the vector, if so, as a function from a set into the field.
The set, of course, is the basis; or some other set with the same number of elements. For a vector $v = \sum_{n=1}^k\alpha_n\cdot v_n$ we can think of $v$ as a function from the set $\{1,\ldots,k\}$ into the field: $v(n)=\alpha_n$. Of course, after changing a basis we "change the function", but this is why vector spaces are isomorphic and not the same.
Best Answer
Every operation is a type of function. Not every function is a type of operation. An "operator" is a type of operation, hence a type of function. But not every operation is an operator.
If $A$ and $B$ are sets, a function $f$ from $A$ to $B$ is a subset of $A\times B$ that satisfies the following two conditions:
If $f$ is a function, and $(a,b)\in f$, we denote this by writing $f(a)=b$. If $f$ is a function from $A$ to $B$, we denote this by writing $f\colon A\to B$.
If $A$ is a set and $n$ is an ordinal (think finite number, but you can do it more generally), then an $n$-ary operation on $A$ is a function $f\colon A^n\to A$. Note in particular that for a function to be an operation, there must exist a set $A$ such that the domain of $f$ is of the form $A^n$ for some ordinal $n$, and the codomain of $f$ is the set $A$.
An operator is a $1$-ary operation, that is a function $f\colon A\to A$. However, that type of nomenclature is usually reserved for specific circumstances, such as a linear function from a vector space to itself rather than in general.
The function mapping $\mathbb{Q}$ to $\mathbb{R}$ by sending $a\in\mathbb{Q}$ to $a\sqrt{2}\in \mathbb{R}$ is a function, but not an operation.
The function mapping $\mathbb{R}^2\to\mathbb{R}$ by sending $(a,b)$ to $a+b$ is a (binary) operation on $\mathbb{R}$, but not an operator on $\mathbb{R}$.
The function $T\colon \mathbb{R}\to\mathbb{R}$ given by $T(x)=2x+1$ is a (unary) operation on $\mathbb{R}$. It is also a function from $\mathbb{R}$ to itself; it is also an operator on $\mathbb{R}$, though we don't usually refer to it that way.