Here is the problem
Solve $y'(t) = 1 - \int_{0}^{t} y(t - v)e^{-2v}dv$
The solution sets $\mathcal{L}(y) = Y(s)$ and does the following
Notice that in step 1, they have $$Y(s)\dfrac{1}{s+2}$$
Are they implying $$\mathcal{L}(y(t) * e^{-2t}) = \mathcal{L}(y(t)) \mathcal{L}(e^{-2t}) = Y(s)\mathcal{L}(e^{-2t}) = Y(s)\dfrac{1}{s+2}$$
Yes they are implying that. Applying the Laplace transform to a convolution gives a product. Write
$$\mathcal{L}\{f*g\}(s):=\int_0^\infty \left(\int_0^x f(u)g(x-u)du\right)e^{-sx}dx=\iint_D f(u)g(x-u)e^{-sx}dudx$$
The region of integration in the $xu$-plane is $D=\{(x,u):0\le u\le x\}$, an infinite triangle. Writing out the substitution $v=x-u$ our region of integration in the $uv$-plane is simply $(0,\infty)\times(0,\infty)$, so
$$=\iint_{(0,\infty)^2} f(u)g(v)e^{-s(u+v)}dudv=\int_0^\infty f(u)e^{-su}du\int_0^\infty g(v)e^{-sv}dv=\mathcal{L}\{f\}(s)\cdot \mathcal{L}\{g\}(s).$$
Note the Jacobian determinant of the transformation $(x,u)\mapsto(u,x-u)$ is simply $1$ (in abs. value).