Vtyjkyuk

We have the following integral in the time domain :
$$int_0^infty e^ucos(2(u-t))y(u)du$$
This is very similar to the convolution integral. I'm not sure if the $2$ scaling in the cosine function causes problems and we can't say that this is convolution. If it is, the convolution is multiplication in the s-domain and the transform we are looking for is :

$$L[e^uy(u)] cdot L[cos(2(u-t))$$ The first part is easily found using the complex shift property giving $Y(s-1)$. Now for the second part we'll have to use the time shift property along with the scaling property. My question is what comes first? Or maybe, is the function shifted by $t$ or $2t$? I always get confused when scaling and delay take place at the same time.

Since $$L[cosat]=fracss^2+a^2$$ one solution could be :
$$L[cos(2(u-t))=e^-sfracss^2+4$$ Is that the way to go?

Are you sure that the upper limit of your integral is $infty$ and not $t$? Because, if it is not $t$, that's not a convolution, for Laplace transforms. I'm assuming that it is $infty$, as written.

You can still write down the Laplace transformation integral,
switch the order of integration, and end up with

$$
F(s) = int_0^infty du , exp(u) , y(u)
int_0^infty dt , exp(-st) cos[2(u-t)]
$$

So the first stage is still $mathcalL[cos(2(u-t))]$,
or $mathcalL[cos(2(t-u))]$ of course, since $cos$ is even. Instead of using the time-shifting formula (which would apply when the function of $t-u$ to be transformed includes a Heaviside step function $H(t-u)$,
i.e. is zero for $t<u$, which is not the case here) I think it's clearest to do it using
$$
cos(2t-2u)=cos 2u cos 2t + sin 2u sin 2t
$$
and use the Laplace transforms of $cos 2t$ and $sin 2t$. This gives
$$
mathcalL[cos(2(t-u))] = fracscos 2u+2sin 2us^2+4
$$
The remaining integrations
$$
F(s) = fracss^2+4 int_0^infty du , cos 2u exp(u) , y(u)
+
frac2s^2+4 int_0^infty du , sin 2u exp(u) , y(u)
$$
don't immediately look like simple transforms of $y(u)$.
Indeed,
the presence of the $exp(u)$ term makes me worry about convergence.
But of course, this basic structure of the integral was there from the start;
the Laplace transformation has not altered it much.
Good luck.