Given a random walk $X_{t \ge 0}$ on $\mathbb{Z}$ starting at $0$ with probabilities $P(n, n + 1) = p$ and $P(n, n - 1) = 1 - p$, let $Y = \min\{X_0, X_1 \dots \}$. What is the probability that $Y = X_0$. I said $P(Y = X_0)$ must satisfy
$$P(Y = X_0) = p(p + (1-p)(P(Y = X_0))$$
so $P(Y = X_0) = \dfrac{p^2}{1 - p + p^2}$
Is this correct?
It is not correct. The right hand side of your equation multiplies the $p$ inside parentheses with $1$. That is equivalent to assuming that after two up steps, the random walk can never fall below $X_0$.
The following is based on Steele's book. Since the answer does not depend on value of $X_0$, assume $X_0=0$. Consider $A>0$ and define $f(k)$ as the probability that the random walk starting from $k$ ($-1 \le k\le A$) reaches $A$ before -1. The recursion equation $f(k)=pf(k+1)+(1-p)f(k-1)$ simplifies to $\Delta f(k)=(1/p-1)f(k-1)$ where $\Delta f(k)=f(k+1)-f(k)$.
So $f(k)=f(-1)+\sum_{j=-1}^{k-1}\Delta f(j)=f(0)\sum_{j=0}^{k}(1/p-1)^j=f(0)\frac{(1/p-1)^{k+1}-1}{(1/p-1)-1}$. Boundary condition $f(A)=1$ yields $f(0)=\frac{(1/p-1)-1}{(1/p-1)^{A+1}-1}$. The required probability is $P(Y=X_0)=\lim_{A\to \infty}f(0)$ equals $0$ if $p<0.5$ and $2-1/p$ if $p>0.5$.