For a 1st order reaction, how do you derive the integrated rate law for a product versus time?
If you are doing this on a molar basis, you have to be careful about assumptions like: $$[A_o] = [A]_t + [P]_t$$
In the same way that you have defined a component balance for the reactant system, you can also do so for the product system. Let's look at the concentration of B in your reaction equation $\ce{aA -&gt; bB + cC}$.

Defining the component balance,
$$\frac{d[B]}{dt}=k[A]$$ 
$$[B]_o = 0, t = 0$$
This is essentially saying that the generation of $B$ is proportional to the inverse depletion of $A$ (as defined in your component balance). We know that from the solution of the A component balance that, $$[A](t) = [A]_oe^{-kt}$$ Plugging in this expression to the differential equation for $[B](t)$ yields,
$$\frac{d[B]}{dt} = k[A]_oe^{-kt}$$ Solving this differential equation will yield the following expression for $[B]$,
$$[B](t)=-[A]_oe^{-kt}+C_1$$
Using our intial condition that $[B]=0, t = 0$ we solve for our constant $C_1$ and end up with,
$$[B](t)=[A]_o(1-e^{-kt})$$
Note that this is similar to the form you had already derived, however... this is only for $B$. Since the kinetic expression will be the same assuming no initial mols of $B$ or $C$ and same stoiciometric coefficients (i.e $b = c$) then we can reasonably assume that $$[P]_t = 2[A]_o(1-e^{-kt})$$