The right-hand sides of $(1)$ and $(2)$ are the same, because
$$\psi_0(1-x)=\psi_0(-x)-x^{-1}.$$
Hence we only need to show that the left-hand sides of $(1)$ and $(2)$ are also the same. That is, for any complex number $x$ with $|x|<1$, we need to show that
$$\lim_{n\to\infty}\left|\sum_{k=2}^{n-2} \zeta(k) \zeta(n-k) x^{k-1}-\sum_{k=2}^\infty\zeta(k)x^{k-1}\right|=0.$$
By the triangle inequality, the expression under the limit does not exceed
$$\sum_{k=2}^{n-2} \zeta(k)\bigl(\zeta(n-k)-1\bigr)|x|^{k-1}+\sum_{k=n-1}^\infty\zeta(k)|x|^{k-1}.$$
It suffices to show that both $k$-sums here tend to zero as $n\to\infty$. This is clear for the second $k$-sum, so we shall focus on the first $k$-sum.
Let us fix an integer $m\geq 2$ and restrict to $n\geq m+3$. We shall use the obvious decomposition
\begin{align*}\sum_{k=2}^{n-2} \zeta(k)\bigl(\zeta(n-k)-1\bigr)|x|^{k-1}=&\sum_{k=2}^m \zeta(k)\bigl(\zeta(n-k)-1\bigr)|x|^{k-1}\\+&\sum_{k=m+1}^{n-2} \zeta(k)\bigl(\zeta(n-k)-1\bigr)|x|^{k-1}.\end{align*}
On the right-hand side, the first $k$-sum tends to zero, because $\zeta(n-k)$ tends to $1$ for each $k\in\{2,3,\dots,m\}$. Therefore,
$$\limsup_{n\to\infty}\sum_{k=2}^{n-2} \zeta(k)\bigl(\zeta(n-k)-1\bigr)|x|^{k-1}\leq\zeta(2)\bigl(\zeta(2)-1\bigr)\sum_{k=m+1}^\infty|x|^{k-1}.$$
Now the right-hand side can be made arbitrarily small by choosing $m$ sufficiently large, while the left-hand side is independent of $m$. So the left-hand side is zero, that is,
$$\lim_{n\to\infty}\sum_{k=2}^{n-2} \zeta(k)\bigl(\zeta(n-k)-1\bigr)|x|^{k-1}=0.$$
The proof is complete.
