Is there an operation that takes $a^b$ and $a^c$, and returns $a^{bc}$? I know that multiplying exponents of the same base will give you that base to the power of the sum of the exponents ($a^b \times a^c = a^{b+c}$), but is there anything that can be done with exponents that will give you some base to the power of the product of those exponents: $f(a^b, a^c) = a^{bc}$?

The function as initially described fails to exist by the definition of a function: for any given input (or input "vector") $x$, there must exist a single output $f(x)$.  Here, the quickest counterexample for any such function $f(a^b,a^c)=a^{bc}$ is to note that $2^4=4^2$ and then any combination of these produces multiple outputs for a single specified input:
$$f(2^4,2^8)=f(4^2,4^3)=f(16,16^\frac 32)=2^{32}\neq 4^6\neq 16^\frac 32$$
or more generally,
$$f(a^{bk},a^{ck})=f((a^k)^b,(a^k)^c)=a^{bck^2}\neq (a^k)^{bc}$$
Essentially, the problem is that with two inputs, our function does not have enough information to give us the result that we want.  We must give the function an additional input, whether by specifying that input as a "parameter" that reduces $f$ to one member of a "family" of functions; or as a "variable" that is part of the input vector into $f$.  Personally, since the $a$ referenced in the question looks a lot like it is intended to be a relatively "fixed" parameter, I would choose to solve the problem this way:
$$f_a(q,r) = a^{\log_a q\cdot\log_a r}$$
where you are declaring the function $f$ to be bound by the base $a$ that you intend to work in.  Now, we would give our function some inputs like $f_a(a^b,a^c)$ and expect to get a value $a^{bc}$.
With the additional comments discussing the possibility of using this function to understand the quantity $e^{i\pi}$ by applying our function to $f_e(e^i,e^\pi)$, we would naively apply our function as
$$f_e(e^i,e^\pi)=e^{\log e^i\cdot \log e^\pi} = e^{i\pi}$$
and arrive at the correct answer, but this process fails to provide further meaning or understanding, and also ignores the possibility that our function may take on other values when supplied with complex inputs outside of what we intend.
