We can build the theory of the Fourier transform up in several steps. The integral in the definition converges by an immediate estimate if the function is in $L^1 (\mathbb{R})$. Since $L^1 (\mathbb{R}) \cap L^2 (\mathbb{R})$ (the set of functions which are both absolutely integrable and square-integrable) is dense in $L^2 (\mathbb{R})$ (i.e. every function in $L^2$ can be approximated of functions in the former space which converges in the $L^2$ norm), and the Fourier transform is a continuous linear operator from $L^1 (\mathbb{R})$ to $L^2 (\mathbb{R})$, we can define the Fourier transform for f in $L^2 (\mathbb{R})$ by approximating $f$ with a sequence of functions $f_n \in L^1 (\mathbb{R}) \cap L^2 (\mathbb{R})$, and defining the Fourier transform of $f$ to be the $L^2$-limit of the Fourier transforms of $f_n$. This idea of taking a function defined on a dense subset and extending it by continuity to a larger set is very useful.
Of course, your example isn't in $L^2 (\mathbb{R})$ either. It turns out we can extend the Fourier transform to an even broader class of functions (actually, to a class that includes things that aren't functions in the traditional sense at all!). This broad class is the set of tempered distributions. We do this by first noting that the Fourier integral is well-defined on the Schwartz class, which is roughly speaking the set of functions which decay to zero at infinity faster than any polynomial (hence these functions are all in $L^1 (\mathbb{R})$. We give the Schwartz class a particular (locally convex) topology, which (in particular) determines which sequences in this space are convergent and which functions defined on the Schwartz class are continuous.
The set of tempered distributions is the dual space to the Schwartz class, which means it is the set of all continuous linear functions which map from the Schwartz class to the real (or complex numbers). An example is the Dirac delta, which satisfies $\delta(f) = f(0)$ for $f$ in the Schwartz class. Distributions are objects which act on functions. Operations on distributions may therefore be characterized by how they affect the way the distribution acts on functions. It turns out that for any tempered distribution $\phi$, we can define the Fourier transform $F \phi$ as the tempered distribution defined by
\begin{equation*}
F \phi (g) = \phi (Fg) \quad \text{ for all } g \text{ in the Schwartz class}.
\end{equation*}
So the Fourier transform of a tempered distribution is the distribution which first applies the Fourier transform to the test function, and then applies the original distribution to the result.
Roughly speaking, all functions with polynomial growth at infinity have corresponding tempered distributions, and hence we can define their Fourier transform. Given such a function $f$, we define the tempered distribution corresponding to $f$ by
\begin{equation*}
\Lambda_f (g) = \int_{-\infty}^\infty f(x) g(x) \, dx \text{ for any } g \text{ in the Schwartz class}.
\end{equation*}
In particular, the function in your example has a corresponding tempered distribution, and so we can define the Fourier transform of that distribution. Notice that the Fourier transform involves the Dirac delta, and hence is not a traditional function, but instead is defined in terms of how it acts on other functions, as we might expect from our definition of the Fourier transform on the Schwartz class.
edit: Responding to your question in another comment, it is correct that we cannot take the Fourier transform of $e^{-t}$, since $e^{-t}$ is not a tempered distribution (it "grows too fast"). There are several ways to prove this. Suppose that $f(t) = e^{-t}$ was a tempered distribution, and take for granted that the rules of derivatives and Fourier transforms work for tempered distributions more or less like you expect they should (they do). Then $e^{-t}$ would (as a distribution) solve the differential equation
\begin{equation*}
f' + f = 0.
\end{equation*}
Taking the Fourier transform, we obtain
\begin{equation*}
(2 \pi i \xi + + 1) \hat{f} (\xi) = 0 \text{ for every } \xi \in \mathbb{R}
\end{equation*}
where $\hat{f}$ is the Fourier transform of $f$. The only way for this equation to be satisfied is if $\hat{f}$ is identically zero. But (it turns out that) the Fourier transform on the set of tempered distributions is an isomorphism, and in particular is invertible, so $\hat{f} = 0$ implies $f = 0$, since the Fourier transform of $0$ is clearly zero. This is a contradiction, so $f$ cannot be a tempered distribution, so we can't define its Fourier transform.
Best Answer
I think the clearest way to see this is by noting that we have (depending on your convention for the placement of $2 \pi$ in Fourier transforms) that $$\mathcal{F}(\mathcal{F}(f(x))) = 2 \pi f(-x)$$ Taking the convention that $$\tilde{f}(k) = \int_{-\infty}^\infty e^{ikx} f(x) \; dx$$ so $$f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx} \tilde{f}(k) \; dk$$ we get $$f(-x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-ikx} \tilde{f}(k) \; dk = \frac{1}{2\pi} \mathcal{F(\tilde{f}(k))} = \frac{1}{2\pi}\mathcal{F}(\mathcal{F}(f(x)))$$ Note we have $$\mathcal{F}(\delta(x)) = \int_{-\infty}^\infty e^{ikx} \delta(x) \; dx = 1$$ So then $$\mathcal{F}(1) = \mathcal{F}(\mathcal{F}(\delta(x))) = 2 \pi \delta(-x) = 2 \pi \delta(x)$$ For other constants, note by linearity we have $$\mathcal{F}(c) = c \mathcal{F}(1) = 2 \pi c \delta(x)$$