Notice that $\sigma(p^{k-1}) = k$ and so the image of $\sigma$ is all of $\mathbb{N}$.

By the way, $\sigma$ is usually used for the sum of divisors function, and it is more standard to use $d$ or $\tau$ for your function.

EDIT: I misread the question. I will use $\tau$ instead of $\sigma$.

I claim that $\mu ^ {+}(\tau^{-1}(\{k\})) = 0$. Take a number $m$ in this set, and let us look at $m$'s prime factorization: $m = p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_r^{\alpha_r}$. Notice that there are finitely many options for the $\alpha_i$ (up to a permutation), because $(\alpha_1 + 1)(\alpha_2 + 1) \cdots (\alpha_r + 1)=k$, so it is enough to show that the upper density of numbers of the form $p_1 ^ {\alpha_1} p_2 ^{\alpha_2} \cdots p_r^{\alpha_r}$ where $r, \alpha_i$ are fixed is zero.

Let us look at numbers in this set that are at most $x$. Then if we fix $p_1$, we need to choose primes $p_2, \cdots p_r$ such that $p_2 ^{\alpha_2} \cdots p_r^{\alpha_r} \leq \frac{x}{p_1 ^{\alpha_1}}$.

By induction we can assume that the amount of numbers of the form $p_2 ^{\alpha_2} \cdots p_r^{\alpha_r}$ which are at most $x$ is $o(x)$, and if $\alpha_1 \geq 2$ then this shows that the amount of numbers of the form $p_1 ^{\alpha_1} p_2 ^{\alpha_2} \cdots p_r^{\alpha_r}$ is $o(x)$ by summing over the options of $p_1$ (and using the fact that $\sum_{p} \frac{1}{p^2} \leq \sum_{n} \frac{1}{n^2}$ converges. Therefore it is enough to solve in this case where all $\alpha_i$ are 1, that is to show that the amount of numbers of the form $p_1 \cdots p_r$ up to $x$ is $o(x)$ (for $r$ fixed).

Fixing $p_1$ we see that $p_2$ can be any prime that is at most $\frac{x}{p_1}$. and then $p_3$ can be anything that is at most $\frac{x}{p_1 p_2}$, ... and $p_r$ is at most $\frac{x}{p_1 p_2 \cdots p_{r-1}}$. So we see that the amount of numbers at most $x$ is

$$\sum_{p_1 \leq x} \sum_{p_2 \leq \frac{x}{p_1}} \cdots \sum_{p_{r-1} \leq \frac{x}{p_1 \cdots p_{r-2}}} \pi (\frac{x}{p_1 \cdots p_{r-1}})$$

From here we can use the simple bound $\pi (x) \leq \frac{cx}{log x}$ for some constant $c$ and see that this sum is small.