Total voltage towards load?
The way you present the circuit in your question is very difficult to follow. Here are the same circuits drawn in a more logical way. It helps to take some time to draw the circuits in a way that's easier for others to interpret.







simulate this circuit &ndash; Schematic created using CircuitLab

Notice that V1 and V2 are not strictly in parallel. The diodes isolate them from each other.

The circuit without a load is a little confusing to solve if you're thinking of ideal components. What happens to ideal voltage sources when you connect them together? Infinities are involved. And it doesn't bring us any closer to understanding what's really happening.

Luckily non-ideal components in the real world makes this a little easier. We know immediately that D1 will not conduct. Whatever the voltage on the cathode of D1 is, it'll be larger that 5V. So we can ignore V2 and D1 in both versions of the circuit.

The anode of D2 is 20V, which is larger than the 15V from V3, so D2 will conduct. Right at the cathode of D2 will be 20V minus the forward drop. That will undoubtedly be more than V3's 15V, so current is gonna flow into V3. In the circuit without a load, one of three things will happen.  

1) If V1 and V3 are "stiff" enough, they both hold their respective voltages. The current out of V1 and through D2 will increase until the natural resistance of the physical wire connecting D2 to V3 provides the necessary voltage drop according to Ohm's Law. So the voltage on that wire will actually be a linear function along its length. Physically closer to V1, the voltage will be closer to 20V. Physically closer to V3 will be closer to 15V.  

2) If V1 is "stiffer" than V3, the current flow into V3 will elevate its voltage to be closer to V1. In that case, the voltage along the wire will be closer to 20V. In a real world battery, this might make V3 explode.  

3) If V3 is "stiffer" than V1, the current draw out of V1 will cause V1's voltage to drop. In that case, the voltage along the wire will be closer to 15V.

The circuit with the 2 Ohm resistor is a little easier. Assuming the voltage sources are adequate, the left side of the resistor will be about 20V minus the diode drop. The right side of the resistor will be 15V.

EDIT:
Based on discussion in the comments, it looks like you're trying to come up with a way to make V1 and V3 cancel each other out so V2 dominates. Since you haven't explained exactly why you're doing this, I'll take an academic approach without regard to practical implications.

Voltage sources will cancel each other out if they're in series and in opposite polarity to each other. And they do not share a common reference voltage of any kind. Therefore, to make V1 cancel V3 and allow V2 to dominate, you would simply do the following. Notice the orientation of V3 relative to V1.







simulate this circuit