Let me complete Sam Hopkins' answer.
The expression on the left is $$ \Psi\left(\sum_{i_1+\dots+i_{n+1}=n-1}{n-1\choose i_1,\dots,i_{n+1}}\prod_j x_j^{i_j+1}\right) =\sum_{i_1+\dots+i_{n+1}=n-1}{n-1\choose i_1,\dots,i_{n+1}}\prod_j(i_j+1)! =(n-1)!\sum_{i_1+\dots+i_{n+1}=n-1}\prod_j (i_j+1) =(n-1)!\sum_{a_1+\dots+a_{n+1}=2n}\prod_j a_j. $$ Now the identity is a partial case of the following: $$ \sum_{a_1+\dots+a_k=\ell}\prod_j a_j={\ell+k-1\choose 2k-1}, $$ as $$ n!\frac1{2n+1}{3n\choose n}=\frac{(3n)!}{(2n+1)!}=(n-1)!{3n\choose 2n+1}. $$
To prove the general identity, denote the left-hand part by $f(k,\ell)$ and notice that $f(1,\ell)=\ell$ and $$ f(k,\ell)=\sum_i if(k-1,\ell-i). $$ Now it suffices to check that the right-hand part satisfies the same conditions, which is easy (the recurrence follows from $(1-x)^{1-2k}=(1-x)^{-2}(1-x)^{3-2k}$).