Let $\tau$ be the first visit 1 by an asymmetric random walk $P(+1)=p>0.5, P(-1) = 1-p.$
By the conditioning on the first step, ($\mathbf{E}^p$ means the random walk started at $p$) \begin{align*} \varphi(z) := \mathbf{E}^0(z^\tau) = (1 - p)z\mathbf{E}^{-1}(z^\tau) + p z. \end{align*} By the strong Markov property applied at the first time to visit zero, $\mathbf{E}^{-1}(z^\tau) = \varphi(z)^2.$ Hence, we reach that $\varphi(z)$ satisfies the "functional equation" \begin{align*} (1-p)z \varphi(z)^2 + pz = \varphi(z). \end{align*}
This equation is trivially true whenever $\varphi(z) = \infty,$ and when $\varphi(z) < \infty$ it can be solved explicitly using quadratic formula. And for those values of $z$ for which $\varphi(z) < \infty,$ this function is explicitly expressible as $\varphi(z) = \dfrac{1-\sqrt{1-4p(1-p)z^2}}{2(1-p)z}$ (the trivial relation $\varphi(0) = 0$ allows deducing minus is the right choice of sign).
It is easy to see that $0 \leq \varphi(z) \leq 1$ whenever $0 \leq z \leq 1.$ Hence the previous explicit expression applies. However, that expression is well-defined for $z \leq \dfrac{1}{2 \sqrt{p(1 - p)}}.$ A bit of algebra shows that $1 < \dfrac{1}{2 \sqrt{p(1 - p)}}$ (since $p > 0.5$).
Question. It is open the possibility that $\varphi(z) < \infty$ for values of $1 < z \leq \dfrac{1}{2 \sqrt{p(1 - p)}}.$ Any idea on how to show that?