Let $f(t)=e^{-j\omega_0 t}$. Note that $f\notin L^1$ and so the object written $\int_{-\infty}^\infty e^{j(\omega-\omega_0)}\,dt$ is not defined as either a Lebesgue integral or an improper Riemann integral .
However, the Fourier Transform of $f$ can be defined in terms of Tempered Distributions. Tempered Distributions are linear continuous mappings from the Schwartz Space $\mathbb{S}$ to the complex numbers. The Dirac Delta is a Tempered Distribution.
Let $\phi\in \mathbb{S}$ (i.e., $\phi$ is a Schwartz Function). Then, the Fourier Transform of $f$ is given distributionally by
$$\begin{align}
\langle \mathscr{F}\{f\},\phi\rangle&=\langle f,\mathscr{F}\{\phi\}\rangle\\\\
&=\int_{-\infty}^\infty e^{-j\omega_0 t}\int_{-\infty}^\infty \phi(\omega)e^{j\omega t}\,d\omega\,dt\\\\
&=2\pi \phi(\omega_0)
\end{align}$$
where the last equality is a consequence of the Fourier Inversion Theorem (See Appendix). Therefore, in distribution we have
$$\mathscr{F}\{f\}(\omega)=2\pi\delta(\omega-\omega_0)$$
APPENDIX: PROOF OF THE INVERSION THEOREM FOR SCHWARTZ FUNCTIONS
In THIS ANSWER, THIS ONE and in the Alternative Development section of THIS ANSWER, I proved that $\frac{\sin(Lt)}{\pi t}$ is a nascent Dirac Delta. Using this result, we see that
$$\begin{align}
\int_{-\infty}^\infty e^{-j\omega_0 t} \int_{-\infty}^\infty \phi(\omega)e^{j\omega t}\,d\omega\,dt&=\int_{-\infty}^\infty \int_{-\infty}^\infty \phi(\omega+\omega_0)e^{i\omega t}\,d\omega\,dt\\\\
&=\lim_{L\to \infty} \int_{-\infty}^\infty \phi(\omega+\omega_0) \int_{-L}^L e^{j\omega t}\,dt\,d\omega\\\\
&=\lim_{L\to \infty} 2\pi \int_{-\infty}^\infty \phi(\omega+\omega_0) \frac{\sin(L\omega)}{\pi \omega}\,d\omega\\\\
&= 2\pi\phi(\omega_0)
\end{align}$$
as expected!
