Is there a constructive discontinuous exponential function? It is well-known that the only continuous functions $f\colon\mathbb R\to\mathbb R^+$ satisfying $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ are the familiar exponential functions.  (Prove $f(x)=f(1)^x$ successively for integers $x$, rationals $x$, and then use continuity to get all reals.)

The usual example to show that the identity $f(x+y)=f(x)f(y)$ alone doesn't characterize the exponentials requires the axiom of choice.  (Define $f$ arbitrarily on the elements of a Hamel basis for $\mathbb R$ over $\mathbb Q$, then extend to satisfy the identity.)

Is there an explicit construction of a discontinuous function satisfying the identity?  On the other hand, does the existence of such a function imply the axiom of choice or some relative?
Let $g(x)=\ln f(x)$, so that $g(x+y)=g(x)+g(y)$. The solutions to this equation are precisely the ring homomorphisms from $\mathbb R\to\mathbb R$ and they are in bijection with solutions to your original equation. Since $\mathbb R$ is an infinite dimensional $\mathbb Q$-vector space, there are infinitely many such ring homomorphisms. They are determined by a Hamel basis for $\mathbb R$.

Does this answer your question?
Let $g(x)=\ln f(x)$, so that $g(x+y)=g(x)+g(y)$. The solutions to this equation are precisely the ring homomorphisms from $\mathbb R\to\mathbb R$ and they are in bijection with solutions to your original equation. Since $\mathbb R$ is an infinite dimensional $\mathbb Q$-vector space, there are infinitely many such ring homomorphisms. They are determined by a Hamel basis for $\mathbb R$.

Does this answer your question?