I tried to solve the laplace transform of $\cos(at)$ and $\sin(at)$ using Euler's formula. That is, $$\int^\infty_0e^{-(s-ia)t}dt\color{red}{=}\frac{1}{s-ia}\int^\infty_0e^{-t}dt=\frac{1}{s-ia}$$ Taking the real and imaginary parts immediately gives $\mathcal{L}\left\{\cos(at)\right\}(s)=\dfrac{s}{s^2+a^2}$ and $\mathcal{L}\left\{\sin(at)\right\}(s)=\dfrac{a}{s^2+a^2}$.
My question is: Is it possible to justify the first equality using contour integration?
You can justify that equality by a substitution. Let $u = (s-ia)t$. Then $du = (s-ia)dt$. $$ \int_0^{\infty}e^{-(s-ia)t}dt = \frac{1}{s-ia}\int_0^{\infty}e^{-u}du $$ Since $u$ is a dummy index, we can let $u = t$ so $$ \frac{1}{s-ia}\int_0^{\infty}e^{-t}dt = \frac{1}{s-ia} $$ If then multiple by $1=\frac{s+ia}{s+ia}$, we have $$ \frac{s+ia}{s^2+a^2} $$ and as you say \begin{align} \text{Re}\biggl\{\frac{s+ia}{s^2+a^2}\biggr\}&= \frac{s}{s^2+a^2}\\ \text{Im}\biggl\{\frac{s+ia}{s^2+a^2}\biggr\}&= \frac{a}{s^2+a^2} \end{align} We can also take the inverse Laplace transform of $\frac{1}{s-ia}$ \begin{align} \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{e^{st}}{s-ia}ds &= \lim_{s\to ia}(s-ia)\frac{e^{st}}{s-ia}\\ &= e^{iat} \end{align} which is what we took the Laplace transform of to begin with, $\mathcal{L}\{e^{iat}\}$.