This is a messy question, because the Fock states over the different bases are different: that is, the state
$$
|n_1,n_2,n_3,\ldots\rangle_\lambda
$$
means a state with $n_1$ particles in state $|\lambda_1\rangle$, $n_2$ particles in state $|\lambda_2\rangle$, and so on, and this is pretty messy if you try to expand all of those $|\lambda_j\rangle$ states into the $|i\rangle$ basis. However, the structure is a bit easier to understand if you go back to the initial definition, as
$$
|\{n\}\rangle_\lambda
=
\frac{1}{\sqrt{N!}}\sum_{P\in S_N}(\pm1)^P
P |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k},
$$
and to the action of $a_{\lambda_1}^+$ on that basis,
$$
a_{\lambda_1}^+
|\{n\}\rangle_\lambda
=
\frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P
P |\lambda_1\rangle^{\otimes (n_1+1)} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}.
$$
Is there something we can do to simplify how this transforms to the $|i\rangle$ basis? Well, the only thing that's changed from the previous state, so one thing we should really try to go for is just the identity
$$
a_{\lambda_1}^+|\{n\}\rangle_\lambda
=
\sum_i \langle i|\lambda_1\rangle a_{i}^+|\{n\}\rangle_\lambda,
$$
which is a lot more reasonable. To put this into action, we just expand that single added state into the new basis:
$$
a_{\lambda_1}^+
|\{n\}\rangle_\lambda
=
\sum_i \langle i|\lambda_1\rangle
\frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k},
$$
so we've done a good deal of the job.
What remains to be done? We want that right-hand side to really match the action of our transformed creation operator,
$$
a_{i}^+|\{n\}\rangle_\lambda
=
\frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k},
$$
and here the easy way out is just to say "hey, we've just put in an extra $|i\rangle$, what else do you need?, which does work but which you might be a bit unhappy with. If you really want to do the grueling details, you need to expand all the states in that tensor product, giving you an extremely messy sum:
\begin{align}
\frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}
\\ & =
\frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle
\left(\sum_{j_1} |i_{j_1}\rangle\langle i_{j_1} |\lambda_1\rangle \right)^{ \otimes n_1}
\left(\sum_{j_2} |i_{j_2}\rangle\langle i_{j_2}|\lambda_2\rangle\right)^{ \otimes n_2}
\\ & \qquad \otimes
\left(\sum_{j_3} |i_{j_3}\rangle\langle i_{j_3}|\lambda_3\rangle\right)^{ \otimes n_3}
\cdots
\left(\sum_{j_k} |i_{j_k}\rangle\langle i_{j_k}|\lambda_k\rangle\right)^{ \otimes n_k},
\end{align}
and if you expand all of those coefficients out you'll run out of ink. However, there's a common thread in that mess, and it's the fact that all the stuff you're adding up is now Fock states over the new basis:
\begin{align}
\frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}
\\ & =
\sum_\text{mess} \prod_{i',\lambda} \langle i' |\lambda\rangle \times
\frac{1}{\sqrt{(N+1)!}}\sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle
|i_{1}\rangle^{ \otimes n_1'}
|i_{2}\rangle^{ \otimes n_2'}
|i_{3}\rangle^{ \otimes n_3'}
\cdots
|i_{k'}\rangle^{ \otimes n'_{k'}},
\end{align}
and in each of those, you've just got $a_i^+$ acting on the new Fock state,
\begin{align}
\frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}
\\ & =
a_i^+
\sum_\text{mess} \prod_{i',\lambda} \langle i' |\lambda\rangle \times
\frac{1}{\sqrt{N!}}\sum_{P\in S_{N+1}}(\pm1)^P P
|i_{1}\rangle^{ \otimes n_1'}
|i_{2}\rangle^{ \otimes n_2'}
|i_{3}\rangle^{ \otimes n_3'}
\cdots
|i_{k'}\rangle^{ \otimes n'_{k'}},
\end{align}
at which stage you can just replace the mess back to how it was, which, because nothing has changed, just collapse back to the product $|\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}$ and therefore to our initial $\lambda$-basis Fock state,
\begin{align}
\frac{1}{\sqrt{(N+1)!}} & \sum_{P\in S_{N+1}}(\pm1)^P
P |i\rangle |\lambda_1\rangle^{\otimes n_1} |\lambda_2\rangle^{\otimes n_2} |\lambda_3\rangle^{\otimes n_3}\cdots|\lambda_k\rangle^{\otimes n_k}
\\ & =
a_i^+ |\{n\}\rangle_\lambda,
\end{align}
which is exactly what we wanted to prove. And that also concludes the bigger proof.
$$\tag{$\square$}$$
To be honest, I've probably been a bit shoddy with signs and constants at a few points during the proof ─ but because of the structural features, it has to work. The details are left as an exercise for the reader ;-).
