Representing Capacitors and Inductors with Initial Conditions for use in a Thévenin Equivalent Circuit
Okay so lets make it a more general setting first. The question you want to ask yourself is, how does one represent the laplace transform of the nth degree derivative operator, including the initial condition.
The first derivative has the following result:
$$\mathcal{L}\left[\frac{dx}{dt}\right] = sX(s) - x(0^-)$$
Higher order derivatives can be attained by the repeated application of the above result.
The general result is:
$$\mathcal{L}\left[\frac{d^nx}{dt^n}\right] = s^nX(s) - \sum_{k=1}^{n} s^{n-k} x^{(k-1)} (0^-)$$
Note in this notation \$ x^{(r)} (0^-) \$ is the \$ r^{th} \$ derivative evaluated at \$ t = 0^- \$. This is the initial condition.
Now for capacitors:
$$ i = C \frac{dv} {dt} $$
In the laplace domain:
$$ I(s) = C (sV(s) - v(0^-))$$
For inductors, its just the dual:
$$ v = L \frac{di} {dt} $$
In the laplace domain:
$$ V(s) = L (sI(s) - i(0^-))$$
You can now solve the above systems using these results.
For example, I have done the capacitor system for you:
$$ I_1 = C_1 (sV_1 - v_1(0^-)) $$
$$ I_2 = C_2(sV_2 - v_2(0^-)) $$
But
$$ I_1 = - I_2 = \frac{V_2 - V_1}{R}$$
Now lets solve for \$ V_1 \$ by elimating the rest, we get:
$$ V_1 = \frac{(sRC_1C_2+C_1)v_1(0^-) + C_2v_2(0^-)} {s^2RC_1C_2+s(C_1+C_2)} $$
To get the final values for the voltage, we use the final value theorem (\$ \lim_{s\to0}sV_1(s) \$) and get:
$$ v(t\to\infty) = \frac{C_1v_1(0^-) +  C_2v_2(0^-)}{C_1 +C_2} $$
This makes sense because it is essentially saying the new voltage across both capacitors will be the total intial charge redistributed to the total capacitance:
$$ v(t\to\infty) = \frac{Q_1(0^-) + Q_2(0^-)}{C_1 +C_2} = \frac{Q_{Total}(0^-)} {C_{Total}} $$
Plug in your values and you get 40V as shown by LTspice:
$$ \frac{1000\cdot 20 + 500\cdot 80} {1500} = 40V $$
EDIT 1:
So let me make it more clear as to why including the step function works, it follows from this expression:
$$ I(s) = C (sV(s) - v(0^-))$$
Invert the equation in terms of \$ V(s) \$:
$$V(s) = \frac{I(s)}{sC} + \frac{v(0^-)}{s}$$
As a circuit, the above equation describes a series connection of a capacitor and a step function!
EDIT 2 :
Also note how this implements the initial condition as a series circuit. You can also implement it instead as a capacitor and a shunt connected constant current source, if you stick to the original equation in terms of current.