Current flow with a potential difference

In CMOS circuits there is dynamic current flow and it is from one of the power rails to the parasitic capacitance on the output, there is no "closed path" but the current still flows.

There is a closed path. The parasitic capacitance allows current to flow (momentarily) when the voltage on one side of it is changing, thus closing the path.

But this doesn't mean it flows with only one connection to the reference node. The power supply needs to connect from the reference node to the Vdd supply of the CMOS circuit, so that a complete circuit is formed, for current to flow.

There will be EHP recombination at this reference point.

Electron-hole recombination only happens in semiconductors. It happens (as a net effect) whenever the product of elecron and hole population densities is greater than the square of the intrinsic population density (i.e. \$np>n_i^2\$). It has nothing to do with whether the semiconductor is at the reference point or not.

Voltage source V2 is not a a source of absolute voltage, it is a source of potential difference. In this case all you can say is that the difference in potential between the two ends of the voltage source (nodes A and B below) is 1V, but you cannot state that the absolute potential of either A or B is zero-volts, or any other value, without further analysis. In other words \$V_B - V_A = 1V\$:

^{simulate this circuit – Schematic created using CircuitLab}

In fact, since there can be no current flowing here (there's no loop for current flow around), then by Ohm's law the potential difference \$V_{R2}\$ across resistor R2 is zero:

$$ V_{R2} = V_B - V_C = I \times R_2 = 0A \times 100\Omega = 0V $$

Remembering that the ground symbol is merely a way to state "this node has potential 0V", potential \$V_B\$ at node B is:

$$ V_B = V_C + V_{R2} = 0V + 0V = 0V $$

This makes the potential \$V_A\$ at node A less than zero:

$$ V_A = V_B - V_2 = 0V - 1V = -1V $$

The simulator agrees:

^{simulate this circuit}

To ensure that two nodes have exactly the same potential, you join them physically:

^{simulate this circuit}

This raises the potential at A from −1V to 0V, and consequently the potential at B must rise by the same amount, since V2 is still producing a 1V potential difference.

To remain consistent, there now has to be a 1V potential difference across R2, and depending on your philosophical bent, that's either because current is now able to flow around the closed loop, causing the resistor to develop \$V_{R2} = I \times R_2 = 1V\$, or because the newly formed potential difference across R2 pushes current \$I = \frac{1V}{100}=10mA\$ through it. Both are equivalent.

From the perspective of a MOSFET gate (node G), which is a tiny capacitor (between gate G and source S), it can only charge if there's a current loop around which current can flow, through the capacitor. Attaching one end of a voltage source to the gate does not alter that gate's potential, since no current flows. If the gate has potential \$V_G=0V\$, the other end of the voltage source simply falls in potential to \$V_A=-1V\$, and no current flows to change those conditions:

^{simulate this circuit}

For \$V_{GS}\$ to increase (to switch the transistor on; 1V probably isn't enough, but I'll stick with values from your example), you must raise gate potential \$V_G\$ with respect to source potential \$V_S\$. This is easiest to do by connecting A to ground also, but you could connect A to any node you know to have 0V potential:

^{simulate this circuit}

Only then can current \$I_G\$ flow, charging the gate capacitance, and raising gate potential \$V_G\$ to +1V.

This same principle applies everywhere in a circuit. There's no such thing as an "absolute" potential, anywhere. There are only potential differences, and those differences define the sate of a system.

Providing a single conductive path between two otherwise unconnected systems (as you did in your example) doesn't change any existing potential difference in either system. Sure, it brings two nodes in the separate systems to the same potential, but it won't change the potential differences present anywhere else. You'll need a second path, to form a loop around which charges can move, and only then can the state of either system change.