It turns out that for a 1D flow all non diagonal terms of T are zero
This is incorrect, a 1D flow is a flow where only one of the components of the velocity exists. That does not mean that the off diagonal terms of the velocity gradient like $\frac{\partial{v_x}}{\partial{y}}$ are zero. This is why your second derivation fails and you end up assuming that the fluid is inviscid without justification
I don't know if your first derivation is correct, but if you follow this: https://en.wikipedia.org/wiki/Navier–Stokes_equations
And use the last equation in the red rectangle right before "incompressible flow", and do every operation correctly, you should be good.
Edit: It doesn't really matter if you just restrict the variables to be only dependent on $x$, you cannot a priori remove terms from the stress tensor without the constitutive equation, because cauchy equations of motion deal with the divergent of the stress tensor, not the actual value of the stresses, which are determined by the constitutive equation. The best you can do with the equations of motion is:
$$ \nabla \cdot T = 0$$
Which means that the stress is overall uniform, and thats it, that doesn't mean there are no shears, or normal stresses, only that they do not vary in each point of the body. A priori there are infinite possible constitutive equations, they don't necessarily follow linear elasticity equations.
Keeping this always in mind, the mistake in your second derivation is assuming there are no deviatoric stresses. The splitting of the stress tensor into hydrostatic and deviatoric goes by:
$$\begin{bmatrix} T_{xx} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy} & T_{yz}\\ T_{zx} & T_{xy} & T_{zz} \\ \end{bmatrix} = \begin{bmatrix} T_{hyd} & 0 & 0 \\ 0 & T_{hyd} & 0\\ 0 & 0 & T_{hyd} \\ \end{bmatrix} + \begin{bmatrix} T_{xx}-T_{hyd} & T_{xy} & T_{xz} \\ T_{yx} & T_{yy}-T_{hyd} & T_{yz}\\ T_{zx} & T_{xy} & T_{zz}-T_{hyd} \\ \end{bmatrix}$$
Given the constitutive equation, if all variables are dependent only on x, then the off diagonal terms are zero, then the result becomes: $$\begin{bmatrix} T_{xx} & 0 & 0 \\ 0 & T_{yy} & 0\\ 0 & 0 & T_{zz} \\ \end{bmatrix} = \begin{bmatrix} T_{hyd} & 0 & 0 \\ 0 & T_{hyd} & 0\\ 0 & 0 & T_{hyd} \\ \end{bmatrix} + \begin{bmatrix} T_{xx}-T_{hyd} & 0 & 0 \\ 0 & T_{yy}-T_{hyd} & 0\\ 0 & 0 & T_{zz}-T_{hyd} \\ \end{bmatrix}$$
So, just because no shear stress exist, that does not mean that no deviatoric stress exist.