Citing CLRS 3ed pp. 94-95:
In the first case, not only must $f(n)$ be smaller than $n^{\log_b a}$, it must be polynomially smaller. That is, $f(n)$ must be asymptotically smaller than $n^{\log_b a}$ by a factor of $n^\epsilon$ for some constant $\epsilon > 0$.
Note that the three cases do not cover all the possibilities for $f(n)$. There is a gap between cases 1 and 2 when $f(n)$ is smaller than $n^{\log_b a}$ but not polynomially smaller.
For your case we have: $$\begin{align} f(n) & = n \log n\\ n^{\log_b a} & = n^2\\ \end{align}$$
We can then show: $$\begin{align} f(n) & = n \log n\\ & < n^{2-\epsilon} & ^{*}\text{for } \epsilon < 1\\ & < n^2 & \text{for } \epsilon > 0\\ \end{align}$$
Then you see we can pick a $\epsilon$ value in the range $(0, 1)$ to get a polynomial difference of at least $n^\epsilon$ because: $$ f(n) < n^{2-\epsilon} < n^2$$
So yes, it is admissible, but your caution is noteworthy, because there are cases where the Master Theorem can not be applied in scenarios similar to this. One example would be the following:
$$T(n) = 2T\left(\frac{n}{2}\right) + \frac{n}{\log n}$$
You see $\frac{n}{\log n}$, while smaller than $n$, is not polynomially smaller, thus we cannot use the Master Theorem.
$^*$: A polylogarithmic function grows more slowly than any positive exponent