1. Given differential equation

$y\frac{dx}{dy}=x\left(\ln x-\ln y+1\right), \qquad x>0,\ y>0$

We need the solution curve passing through $(x,y)=(e,1)$.

2. Simplify the logarithmic expression

Using $\ln x-\ln y=\ln\frac{x}{y},$ we get

$y\frac{dx}{dy}=x\left(\ln\frac{x}{y}+1\right).$

Divide by $x$:

$\frac{y}{x}\frac{dx}{dy}=\ln\frac{x}{y}+1.$

Now notice that a substitution involving $\dfrac{x}{y}$ will help.

3. Use substitution

Let

$u=\ln\frac{x}{y}=\ln x-\ln y.$

Differentiate with respect to $y$:

$\frac{du}{dy}=\frac{1}{x}\frac{dx}{dy}-\frac{1}{y}.$

Multiply by $y$:

$y\frac{du}{dy}=\frac{y}{x}\frac{dx}{dy}-1.$

From the differential equation,

$\frac{y}{x}\frac{dx}{dy}=u+1.$

So,

$y\frac{du}{dy}=(u+1)-1=u.$

Thus the equation reduces to

$y\frac{du}{dy}=u.$

4. Solve the separable equation

$y\frac{du}{dy}=u \quad \Rightarrow \quad \frac{du}{u}=\frac{dy}{y}.$

Integrating,

$\ln|u|=\ln y + C.$

Hence,

$|u|=Cy,$ where $C>0$ is a constant.

Substitute back $u=\ln\dfrac{x}{y}$:

$\left|\ln\frac{x}{y}\right|=Cy.$

5. Use the point $(e,1)$

At $x=e, y=1$,

$\left|\ln\frac{e}{1}\right|=C(1).$

Since $\ln e=1$,

$|1|=C \Rightarrow C=1.$

Therefore,

$\boxed{\left|\ln\frac{x}{y}\right|=y}.$

6. Match with options

This is exactly Option C:

$\boxed{\left|\log_e\frac{x}{y}\right|=y}.$

So the correct answer is C.