Prove the existence of Laplace transform of $2te^{t^2}\cos(e^{t^2})$

Note that $f(t)=\frac{d}{dt}\sin(e^{t^2})=2te^{t^2}\cos(e^{t^2})$. Then, we have

$$\begin{align} \int_0^R f(t)e^{-st}\,dt&=\int_0^R e^{-st}\frac{d}{dt}\sin(e^{t^2})\,dt\\\\ &e^{-sR}\sin(e^{R^2})-\sin(1)+s\int_0^R \sin(e^{t^2})e^{-st}\,dt \end{align}$$

The function $\sin(e^{t^2})e^{-st}\in C[0,R]$ and is therefore integrable on $[0,R]$ for $\text{Re}(s)>0$. Moreover, we can write for $\text{Re}(s)>0$

$$\begin{align} \left|\int_0^R \sin(e^{t^2})e^{-st}\,dt\right|&\le \int_0^R e^{-st}\,dt\\\\ &=\frac{1-e^{-sR}}{r}\to \frac1s \end{align}$$

Hence for $\text{Re}(s)>0$, $\int_0^\infty \sin(e^{t^2})e^{-st}\,dt$ exists as an improper Riemann integral and is finite and $\int_0^\infty 2te^{t^2}\cos(e^{t^2})e^{-st}\,dt$ also exists and is finite as an improper Riemann integral.

This is an answer to the original question about the Laplace transform of $\sin(e^{t^2})$.
I recommend the OP to try avoiding chamaleon questions in the future and simply ask a new question.

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For any $s>0$ the integral

$$ (\mathcal{L}f)(s)=\int_{0}^{+\infty}\sin(e^{t^2})e^{-st}\,dt $$ is finite since $\left|\sin(r)\right|\leq 1$ for any $r\in\mathbb{R}$ and $\int_{0}^{+\infty}e^{-st}\,dt = \frac{1}{s}$ is finite.
We have $$ (\mathcal{L}f)(s)=\frac{1}{2}\int_{0}^{+\infty}\frac{\sin(e^t)}{\sqrt{t}}e^{-s\sqrt{t}}\,dt = \frac{1}{2}\int_{1}^{+\infty}\frac{\sin(u)\,du}{ u\,e^{s\sqrt{\log u}}\sqrt{\log u}}$$ that is convergent by Dirichlet's test too, since $\sin u$ has a bounded primitive while $u\,e^{s\sqrt{\log u}}\sqrt{\log u}$ is an increasing function on $(1,+\infty)$.

Notice that

$$2te^{t^2}\cos(e^{t^2}) = \dfrac{d}{dt} \sin(e^{t^2})$$

And that if $$\mathscr{L}\{f(t)\} = F(s)$$ then $$\mathscr{L}\{f'(t)\} = sF(s) -f(0)$$

and also maybe apply @Mark_Viola's rationale for why F(s) must exist.