The question I had was
Find the Laplace transform of $$f(t)=10e^{-200t}u(t).$$
Would it be correct to take out the $10$ because it is a constant, find the Laplace transform of $e^{-200t}$ and then multiply it by the Laplace transform of $u(t)$ to obtain a final answer of $$10\left(\frac{1}{s+200}\right)\left(\frac1s\right).$$ The $u(t)$ is what is really confusing me in this problem.
On
Yes, and no.
It's OK to break out the 10 since the Laplace transform is linear, that is fox $\mathcal L\alpha f = \alpha\mathcal Lf$ for any constant $\alpha$.
But to separately transform $e^{-200t}$ and $u(t)$ and multiply the results is wrong. Multiplication under the $\mathcal L$ corresponds to convolution outside it.
However if you're talking about the one-sided laplace transform then the transform of $e^{-200t}u(t)$ is the same as for $e^{-200t}$ since we only integrate over positive numbers anyway. The result would consequently be:
$$\mathcal L 10e^{-200t}u(t) = 10\mathcal Le^{-200t} = 10 {1\over s+200}$$
There is an extra $1/s$ in your final formula. Let $f(t) = a e^{-b t} u(t)$ with $a,b \in \mathbb{R}$ (I'm assuming your $u(t)$ is the Heaviside step function). Then
\begin{align} (L f)(s) &= \int_0^\infty f(t) e^{-s t} \mathrm dt = \int_0^\infty a e^{-b t}e^{- st} u(t)\mathrm dt \\ &= a \int_0^\infty e^{-(b+s)t} \mathrm dt = \frac{a}{s+b}. \end{align} The region of convergence is $s \in \mathbb{C}: \operatorname{Re}(s) > -b$.