Is it possible to make sense, in distributional sense, of the Fourier transform of the exponential function (defined over the whole real line)?

It is possible to make sense of this if one generalizes the notion of distribution by choosing a smaller space of test functions. The space of test functions should be chosen so that it is closed under Fourier transform and its elements decrease so fast that multiplying them by an exponential function is still integrable. Some possibilities for the space of test functions with these properties are: *Holomorphic functions on the complex plane that decrease faster than any exponential on horizontal strips (if I have remembered the PaleyWiener theorem correctly...) *A more extreme space of test functions is polynomials times Gaussians. (Using smooth compactly supported test functions as suggested in another answer does not work as this space is not closed under Fourier transforms) 


As commented above, $e^{tx}$ with t real is not a tempered distribution, so the simplest standard extension of Fourier transform to tempered distributions does not apply. However, also as commented above, shrinking the space of Schwartz functions to something smaller, the extreme case being compactlysupported smooth, makes real exponentials continuous functionals, so Fourier transform can be defined. True, since Fourier transform does not map test functions to themselves (but to PaleyWiener spaces), such extensions of Fourier transforms will map real exponentials to elements of the dual of such veryslightlyexotic spaces, rather than to spaces of distributions. I think this is a natural outcome. 


The Fourier transforms of distributions (not necessarily of Schwartz class) form a space called analytic functionals. A detailed discussion of these can be found in Gelfand and Shilov. 


L. Ehrenpreis, in his series of papers on the division problem in the 1950's and in his book Fourier analysis in the complex domain, introduced and developed the Fourier transform of distributions. A Fourier transform of a distribution $u\in C^{\infty}(\Omega)$, $\Omega$ an open, convex subset of euclidean $n$space, is a Radon measure $\mu$ such that $$u(\varphi)=\int \hat\varphi(z)\,d\mu(z),\quad \varphi\in C_c^\infty(\Omega),$$ where $\hat\varphi$ is the FourierLaplace transform of the test function $\varphi$. The measure $\mu$ satisfies uniform bounds which derive from the PaleyWiener estimates. (The PaleyWiener theorem has been alluded to in some answers.) More precicely, Ehrenpreis developed his theory of analytically uniform (AU) spaces for the purpose of Fourier representation of general distributions. "The" Fourier transform $\mu$ of $u$ is not unique of course. Still, it is very useful in proving, e.g., the EhrenpreisMalgrange theorem and the Ehrenpreis Fundamental Principle about exponential representation of solutions to homogeneous linear partial differential system with constant coefficients. The Dirac measure at a complex number $\lambda$ is, of course, a Fourier transform of $u(t)=e^{i\lambda t}$. 


Of course, yes. Given a real number $\lambda$ define the distribution $$ E_{\lambda}(u) = \int e^{i \lambda x} u(x) dx $$ for all test functions $u$ (that is $u$ is smooth and compactly supported). Then the Fourier transform of $E_{\lambda}(u)$ is defined by $$ \hat{E_{\lambda}}(u) = E_{\lambda}(\hat{u}) $$ which you can check to be equal to $\hat{u }(\lambda)$. The interpretation is that you extend the equality $$ \int f(x) \hat{g}(x) dx = \int \hat{f}(x) g(x) dx $$ valid for all test functions $f,g$ to distributions. 

