I have found a proof. I hope someone else can give a more conceptual argument.
Let $\Psi_{\lambda}(x)=L_{\lambda}(x)_n$. We approach the expansion of $\Psi$ differently. Begin with \begin{align} \prod_{i=1}^k(E^{\lambda_i}-1) &=\sum_{T\subset\lambda}(-1)^{k-\#T}E^{\vert T\vert} \\ &=\frac12\left(\sum_{T\subset\lambda}(-1)^{k-\#T}E^{\vert T\vert}+ \sum_{T^c\subset\lambda}(-1)^{k-\#T^c}E^{\vert T^c\vert}\right)\\ &=\frac12\sum_{T\subset\lambda}(-1)^{\#T}\left( (-1)^kE^{\vert T\vert}+E^{n-\vert T\vert}\right);\end{align} where $\#T=$ the cardinality of $T$ (if empty then zero), $\vert T\vert=$ sum of elements of $T$ and $T^c$ is the complement of $T$ in the set $\lambda$.
The next step uses a couple of key facts, namely: $$(x+n-q)_{n+1}=(-1)^{n+1}(-x+q)_{n+1} \qquad \text{and} \qquad \frac1{\delta}(x)_n=\frac{(x)_{n+1}}{n+1}.$$ We thus compute \begin{align} \Psi_{\lambda}(x)&=\frac1{\delta}\prod_{i=1}^k(E^{\lambda_i}-1)(x)_n\\ &=\frac1{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^k (x+\vert T\vert)_{n+1}+(x+n-\vert T\vert)_{n+1}\right)\\ &=\frac1{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^k (x+\vert T\vert)_{n+1}+(-1)^{n+1}(-x+\vert T\vert)_{n+1}\right)\\ &=\frac{(-1)^{n+k+1}}{2(n+1)}\sum_{T\subset\lambda}(-1)^{\#T}\left((-1)^{n+1} (x+\vert T\vert)_{n+1}+(-1)^k(-x+\vert T\vert)_{n+1}\right)\\ &=(-1)^{n+k+1}\Psi_{\lambda}(-x).\end{align} The proof is complete. $\square$