Not an answer (yet?).
At first, let's count all the other partitions onto $m$ parts. These are partitions for which $k$'s largest part never equals $k$, $k=1,2,\dots$. We want to prove that the generating function equals $$ x^m\sum_{i=0}^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}} $$ (I have used that total number of partitions onto $m$ parts have generating function $x^m/(x)_m$, that is seen by duality.) Call such partitions interesting.
Multiply by $t^m$ and sum up by $m$. We get a double generating function $f(t,x)=\sum t^{{\rm parts}(\lambda)} x^{|\lambda|}$, summation is taken over all interesting partitions $\lambda$. We have to prove that $$ f(t,x)=\sum_{m\geqslant i\geqslant 0} (tx)^m\frac{(-1)^{i}x^{i(i-1)/2}}{(x)_{m-i}}= \left(\sum_{k\geqslant 0} \frac{(tx)^k}{(x)_k}\right)\left(\sum_{i\geqslant 0} (-1)^i(tx)^i x^{i(i-1)/2}\right). $$ As for the first multiple, it is a double generating function for $t^{\max(\lambda)} x^{|\lambda|}$ taken by all partitions $\lambda$, hence by duality it is the same thing as a double generating function for $t^{{\rm parts}(\lambda)} x^{|\lambda|}$, which is $\prod_{k\geqslant 1} (1-tx^k)^{-1}$. As for the second multiple, it is a part of Jacobi triple product $$ \prod_{n\geqslant 1} (1-tx^{n})(1-t^{-1}x^{n-1})(1-x^n)=\sum_{i=-\infty}^{\infty} (-1)^it^i x^{i(i+1)/2},$$ corresponding to non-negative $i$.
Now we have to do something. I think, we may try bijective proof of Jacobi's triple product with 'particles' and study what exactly corresponds to non-negative $i$.