Question: For example:

$y' = y + y^2$

The Laplace of the first two terms is $s(F(s)-f(0))$ and $F(s)$. 

But what is the Laplace of $y^2$?

Answer: 
As Ian has pointed out, it is hard to deal with the laplace transform of the equation. You can solve the equation in this way:

$$\frac{y'}y =y+1$$

$$\ln (|y|)=\int y\,dx +x $$

Again

$$\frac{y'}{y+1}=y$$

$$\ln(|y+1|) = \int y \,dx$$

So
by taking the difference:

$$\ln \left( |\frac y{y+1}| \right) = x+C$$

$$\frac y{y+1} =\pm k e^x$$

$$y=\frac 1{1 \mp ke^x} -1= \frac{\pm ke^x}{1 \mp ke^x}$$
Thanks guys for the help. The partial fraction method makes the most sense. 

I found out that what I posted was a particular case of Bernoulli equation, which you might find interesting.

<http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx>

We can make the substitution

$$v=1/y$$

Then

$$\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx} = -v^{-2}\frac{dv}{dx}$$

The original equation becomes 

$$-v^{-2}\frac{dv}{dx} = v^{-1} + v^{-2}$$

Divide by ($-v^{-2}$), multiply by $e^x$ and integrate,

$$e^xv' + e^xv = -e^x$$
$$\frac{d}{dx}\big[e^xv\big] = -e^x$$
$$v = -e^{-x}\int e^xdx$$
$$v = -1 + ce^{-x}$$

Substitute back to $y$

$$y = \frac{1}{ce^{-x}-1} = \frac{e^x}{c-e^x}$$