I believe that it is quite instructive to derive the above generating function using species, in spite of it being very simple, as it showcases some standard tricks in the manipulation of cycle indices.
Observe that the species $\mathcal{Q}$ that represents partitions with unique constitutents is simply $\mathcal{Q} = \mathfrak{P}\left(\sum_{k\ge 1} \mathcal{Z}^k\right).$
Now recall the recurrence by Lovasz for the cycle index $Z(P_n)$ of the set operator $\mathfrak{P}_{=n}$ on $n$ slots, which is $Z(P_n) = \frac{1}{n} \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l}) \quad\text{where}\quad Z(P_0) = 1.$
Let $F_n(z) = Z(P_n)\left(\frac{z}{1-z}\right)$ where the second parenthesis on the right represents cycle index substitution and introduce the generating function $G(y) = \sum_{n\ge 0} F_n(z) y^n,$ so that we are interested in $G(1).$
The recurrence yields $n Z(P_n) = \sum_{l=1}^n (-1)^{l-1} a_l Z(P_{n-l}).$
Substitute for $a_l$, multiply by $y^{n-1}$ and sum over $n\ge 1$ to get $G'(y) = \sum_{n\ge 1} y^{n-1} \sum_{l=1}^n (-1)^{l-1} \frac{z^l}{1-z^l} F_{n-l}(z) \\= \sum_{l\ge 1} (-1)^{l-1} \frac{z^l}{1-z^l} \sum_{n\ge l} y^{n-1} F_{n-l}(z).$ This is $\sum_{l\ge 1} (-1)^{l-1} \frac{z^l}{1-z^l} y^{l-1} \sum_{n\ge l} y^{n-l} F_{n-l}(z) = G(y) \sum_{l\ge 1} (-1)^{l-1} \frac{z^l}{1-z^l} y^{l-1}.$ Therefore $(\log G(y))' = \sum_{l\ge 1} (-1)^{l-1} \frac{z^l}{1-z^l} y^{l-1}.$ Integrating we have $\log G(y) = C + \sum_{l\ge 1} (-1)^{l-1} \frac{z^l}{1-z^l} \frac{y^l}{l} = C + \sum_{k\ge 1} \sum_{l\ge 1} (-1)^{l-1} z^{kl} \frac{y^l}{l} \\= C - \sum_{k\ge 1} \log \frac{1}{1+yz^k}.$ The conclusion is that $G(y) = e^C \exp\left( - \sum_{k\ge 1} \log \frac{1}{1+yz^k} \right) = e^C \exp\left( \sum_{k\ge 1} \log(1+yz^k) \right) \\= e^C \prod_{k\ge 1} (1+yz^k).$
Now $G(0)=1$ and hence the constant obeys $e^C=1$, for a final answer of $G(y) = \prod_{k\ge 1} (1+yz^k).$
In particular the generating function of partitions into unique parts is $G(1) = \prod_{k\ge 1} (1+z^k),$ precisely as it ought to be.