No, first let's observe that we must have $0<x$, from the original equation. When we divided through by $x\ln(x)$, we must note that we cannot have $x=0$ (already taken care of by the domain), and we cannot have $x=1$, so we are eliminating the trivial solution $y=e$. We may find we can include this back into the general solution.
Now, with that out of the way we note:
$$\mu(x)=\exp\left(\frac{1}{u}\,du\right)=e^{\ln(u)}=u=\ln(x)$$
Thus, our ODE becomes:
$$\ln(x)\d{y}{x}+\frac{1}{x}y=e^x$$
If we computed the integrating factor correctly, we should not find that the left side of the equation is:
$$\frac{d}{dx}\left(\mu(x)\cdot y\right)$$
Is this what we have?