Contradiction between the first law of thermodynamics and combined gas law

Part 1

The ideal gas law and first law are independent laws. However, the combination of the two laws shows that the change in internal energy of an ideal gas depends only on charge in temperature, as shown in part 2 below.

Part 2

For the version of the first law that you are using when the gas does work (expands) work is negative (reduces internal energy). So for positive $Delta V$, the first law For a constant pressure process is

$$Delta U =Q-PDelta V$$

From the ideal gas law

$$Delta V=frac{RDelta T}{P}$$

Substituting into first law

$$Delta U =Q-RDelta T$$

Also for constant pressure $Q=C_{p}Delta T$, so

$$Delta U =C_{p}Delta T-RDelta T$$

Finally for an ideal gas

$$R=C_{p}-C_v$$

Substituting in first law

$$Delta U =C_{v}Delta T$$

Which shows internal energy of an ideal gas depends only on the temperature.

Part 3

The molecular level explanation for the last equation is for an ideal gas there are no intermolecular forces (no internal potential energy) so the internal energy is strictly kinetic, which is why the internal energy of an ideal gas depends only on the temperature.

So to clarify, the first law of thermodynamics simply tells us that the internal energy of a system is equal to the thermal energy transferred to the system minus the work done by the system;

Essentially, yes. But I would rather word it that the change in internal energy is equal to the net heat added to the system minus the net work done by the system.

it does not show us how internal energy (and hence temperature) relates to pressure and volume (or more specifically a definite relationship that does not involve accounting for the transfer of T.E.).

You are correct it does not show us how internal energy (and hence temperature) directly relates to pressure and volume without accounting for heat transfer (or, as you call it, transfer of thermal energy, T.E.), unless you combine it with the applicable equation of state (e.g., ideal gas equation). But you don't need the ideal gas equation to determine the change in internal energy.

In the example being considered (a reversible constant pressure expansion process) we know that the heat transfer into the system is $C_{p}Delta T$ and the work done by the system is $-PDelta V$ for $Delta V>0$. You don't need the ideal gas law to calculate the change in internal energy. You do need the ideal gas law to calculate the change in internal energy based only on temperature change, as noted in Part 2, namely, $Delta U=C_{v}Delta T$. Either calculation gives the same result. But, and I need to emphasize this, $Delta U=C_{v}Delta T$ only applies to an ideal gas because it is derived from the ideal gas equation.

Whereas the combined gas law takes it further by showing how temperature, and hence internal energy, affects pressure and volume (and vice versa).

Not sure what you mean by the "combined" gas law. But if you mean the combination of the ideal gas law and the first law, it provides the direct direct relationship between internal energy and temperature, and hence pressure and volume by the gas law.

first law of thermodynamics relates the change in internal energy of a system to total heat and work done (hence pressure and change in volume), whereas the ideal gas law relates temperature to pressure and volume (and by extension, to internal energy too).

Yes, but only the ideal gas law relates temperature to pressure and volume and by extension to internal energy too. It would not apply to a real gas. It is important to understand that an equation of state only provides the relationship between pressure, volume and temperature for different equilibrium states. For example, in a closed system (constant mass) the relationship between states 1 and 2 is

$$frac{P_{1}V_{1}}{T_{1}}=frac{P_{2}V_{2}}{T_{2}}$$

This relationship says nothing about the process or path between the two states. For example, if $P_{1}=P_{2}$ that does not necessarily mean the two states are connected by a constant pressure process. There are many possible paths for which the final and initial pressure is the same. So you can't necessarily relate this to heat, $Q$ by $Q=C_{p}Delta T$ since heat is path dependent and $C_{p}$ only applies to a constant pressure path. The final and initial pressures being the same only means $frac{V_{1}}{T_{1}}=frac{V_{2}}{T_{2}}$.

Hope this helps