For your edited questions:
We impose the condition
$$[x_i, p_j]=i\delta_{ij},\ \ E\leftrightarrow i\frac{d}{dt}$$
to quantize classical system. Here, the important point is $p_i$ is a canonical momentum of $x_i$. When we consider the Newtonian Lagrangean
$$H=\frac{mv_i^2}{2}+V(x),$$ the canonical momentum is $p_i^{NR}=mv_i$, and $[x_i, p^{NR}_j]=i\delta_{ij}$ is the quantizating condition. Thus we can identify $p^{NR}_i\leftrightarrow \frac{1}{i}\frac{\partial}{\partial x_i}$.
On the other hand, in the case of relativistic mechanics, its canonical momentum is
$$p_i^R=m\gamma v_i,$$
so we have to impose the condition
$$[x_i, p^{R}_j]=i\delta_{ij}.$$
In this case, there is a correspondence $p_i^R\leftrightarrow \frac{1}{i}\frac{\partial}{\partial x_i} .$
The important point is that, following the canonical quantization formulation, the conjugate momentum is always interpreted as the derivative of the coordinates. Replacing the Newtonian momentum with a coordinate derivative is not proper quantization.
At this point, let’s revisit the quantization procedure for the energy $E\leftrightarrow i\frac{d}{dt}$. What is important in relativistic theory is that the theory becomes covariant under the Lorentz transformation. By no means does every time variable “$t$” that appears in relativity have to be replaced by proper time “$\tau$”. When considering the equations of motion, the time variables are replaced by proper time because otherwise the theory would not be covariant.
With these in mind, let's revisit the quantization law of energy. We already know the quantizing procedure for momentum:
$$p_i^R\leftrightarrow \frac{1}{i}\frac{\partial}{\partial x_i}. $$
On the other hand, four momentum must be covariant under the Lorentz transformation,
$$p^\mu=\begin{pmatrix}
E \\
p_i^R
\end{pmatrix}
\overset{quantize}{\longrightarrow} \begin{pmatrix}
? \\
-i\partial/\partial x_i
\end{pmatrix} $$
so under the appropriate quantization formulation, the above (?) term and $\partial/\partial x_i $ must form Lorentz covariant four vectors. (Because the left-hand side, $p^\mu$, is Lorentz covariant.)
We already know that $$\partial^\mu =\begin{pmatrix}
d/dt \\
-\partial/\partial x_i
\end{pmatrix} $$
is Lorentz covariant, so the usual quantizing procedure $E\leftrightarrow i\frac{d}{dt}$ is appropriate even in the sense of covariance. In short, relativistic quantizing procedure is summarized by following formula:
$$p^\mu\leftrightarrow i\partial^\mu.$$
Now, consider what would have happened if we had taken
$$E\leftrightarrow i\frac{d}{d\tau}$$as the quantization rule. In this case, the right hand side of
$$p^\mu=\begin{pmatrix}
E \\
p_i^R
\end{pmatrix}
\overset{???}{\longrightarrow} \begin{pmatrix}
i d/d\tau \\
-i\partial/\partial x_i\end{pmatrix}$$
does not preserve Lorentz covariance, so the covariance of the theory is lost. In this sense, too, the quantum counterpart of energy $E$ must be $id/dt$. It is never $id/d\tau$.
Finally, it should be pointed out that the energy $E$ is not Lorentz invariant quantity so there is no need to be $E\leftrightarrow i\frac{d}{d\tau}$.
