Strictly speaking, the circuit has 5 nodes, at points labelled a, b, c, d and f. If you use the modified nodal analysis to solve the circuit, you'll apply KCL at all nodes but one (usualy the reference one) to end up with a system of lineal independent equations. So, you'll apply KCL at nodes a, b, c and d to determine the (initially) unknown voltages at these nodes.
However, this is modified by the presence of voltage sources connected to ground at nodes a and c. Due to the presence of these voltage sources, the voltages at nodes a and c are not unknowns, so you have only 2 "real" unknowns (voltages at b and d), and you have to write only 2 KCL. If you have to choose, you would like to write KCL at b and d, because writing the KCL at a and c involves the currents flowing through the voltage sources, and you don't know how to express these currents in terms of the node voltages, so you avoid writing KCL equations at nodes having voltage sources connected to ground.
So finally you have to write the KCL at b and d to solve the circuit, and you can also get rid of the KCL at d if you add the impedances of the resistor and the capacitor together to have 10-10j Ohm.
So your first equation (KCL) is ok, and all you have to do from that is to express the currents in terms of node voltages and impedances:
$$I_1+I_2−I_3−I_4=0$$
$$I_1 = \frac{E_1-V_b}{-10j}$$
$$I_2 = \frac{E_2-V_b}{+20j}$$
$$I_3 = \frac{V_b}{10-10j}$$
$$I_4 = \frac{V_b}{10}$$
And if you substitute these currents in the first KCL and solve for Vb you obtain:
$$V_b = \frac{1-j}{1-3j} * (2E_1-E_2) = \frac{1}{2-j} * (2E_1-E_2)$$
Knowing Vb (and E1 and E2) allows you to easily determine all other circuit variable.
Well, figured it out.
for Thévenin's theorem to work, it is needed to ignore the capacitor (the load).
And then, the solution is simple.
shortning the capacitor makes it clear the the current and the voltage for Thévenin's theorem is zero, after that it is easy to calculate the resistance with a testing voltage source.
Best Answer
Since you are "familiar with nodal analysis," I'll approach your question from that perspective. Let's draw this up two ways:
simulate this circuit – Schematic created using CircuitLab
(We'll assume that you can specify values for the three resistors and the two voltage sources.)
In the left schematic, the values of \$V_A=V_1\$ and \$V_B=V_2\$ are known, since you have them already referenced from the ground reference, directly. So the solution for \$V_X\$ using KCL (nodal analysis here) is:
$$\frac{V_X}{R_1}+\frac{V_X}{R_2}+\frac{V_X}{R_3}=\frac{V_A}{R_1}+\frac{V_B}{R_2}+\frac{0\:\text{V}}{R_3}$$
Note the symmetry here. The voltage at \$V_X\$ drives a current outward through each of the three resistors. But this must equal the current inward arriving through each of those three resistors from the other sources. But you are probably more familiar with this:
$$\frac{V_X-V_A}{R_1}+\frac{V_X-V_B}{R_2}+\frac{V_X-0\:\text{V}}{R_3}=0\:\text{A}$$
Note that there is no difference. These two above equations are identical. And they solve out as:
$$V_X=R_3\:\frac{R_1\:V_2 + R_2\:V_1}{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}$$
But you could also have solved this as a resulting central voltage, given voltage sources with series impedances. The general equation is:
$$V_X=\frac{\sum_{i=1}^N \left[V_i\cdot \prod_{j=1,j\ne i}^N R_j\right]}{\sum_{i=1}^N\left[\prod_{j=1,j\ne i}^N R_j\right]}$$
In this case, that would turn into:
$$V_X=\frac{V_1\:R_2\:R_3 + V_2\:R_1\:R_3 + 0\:\text{V}\:R_1\:R_2 }{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}=R_3\:\frac{R_1\:V_2 + R_2\:V_1}{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}$$
Same result.
Mostly, I just want to point out that there are lots of ways to approach the left side schematic. Keep your mind flexible. Look at things in different ways and find the equivalent approaches.
To the right side schematic now. Nodal analysis would make you write two equations now, instead of one. \$V_B=V_2\$, so we don't need to worry about node \$V_B\$. But we do now have to worry about a new unknown, \$V_A\$. I have to choose a direction for the current in \$V_1\$ and I'm going to say that it is flowing from (-) to (+), or that it is pointing inward relative to \$V_X\$. So:
$$\begin{align*}\frac{V_X}{R_2}+\frac{V_X}{R_3}&=I_{V_1}+\frac{V_B}{R_2}+\frac{0\:\text{V}}{R_3}\\\\\frac{V_A}{R_1}+I_{V_1}&=\frac{0\:\text{V}}{R_1}\end{align*}$$
But \$V_A=V_X-V_1\$, so:
$$\begin{align*}\frac{V_X}{R_2}+\frac{V_X}{R_3}&=I_{V_1}+\frac{V_B}{R_2}+\frac{0\:\text{V}}{R_3}\\\\\frac{V_X-V_1}{R_1}+I_{V_1}&=\frac{0\:\text{V}}{R_1}\end{align*}$$
Solving these simultaneously (or solving the 2nd equation for \$I_{V_1}\$ and substituting that into the 1st equation) will provide the same answer as before:
$$V_X=R_3\:\frac{R_1\:V_2 + R_2\:V_1}{R_1\:R_2 + R_1\:R_3 + R_2\:R_3}$$
The difference here is that you had to create a variable for the current (unknown, just yet) of \$V_1\$ and you had to recognize that there is a very direct relationship then between \$V_X\$ and \$V_A\$ that is created by the fact that there is a voltage source holding them apart by a fixed value.
It's an extra step. But the answer is the same.
The actual circuit is the same complexity, regardless. But being allowed to call a node "ground" and assign it the special value of \$0\:\text{V}\$ provides one way to help simplify the equations a little. And sometimes, if you move things around (without materially changing the circuit topology) you can simplify them a little more. But regardless of how you approach it, if the question being asked isn't changed and if the circuit remains equivalent, the same results should be obtained.