$\def\Z{\Bbb Z}$This is related to Weibel Exercise 3.3.1 Show that $\operatorname{Ext}_\Z^1(\Z[1/p],\Z)\cong\hat{\Z}_p/\Z\cong \Z_{p^\infty}$.
I have done the problem by the following way. Take the short exact sequence of the followings.
$0\to \Z\to \Z[1/p]\to \Z_{p^\infty}\to 0$ $(\star_0)$
Use projective resolution as the following.
$0\to \Z\to \Z\to 0$
$0\to\oplus_i(p^i\Z)\to \Z\oplus[\oplus_i(\Z)]\to \Z[1/p]\to 0$ $(\star_2)$
$0\to \oplus_i(p^i\Z)\to \oplus_i \Z\to \Z_{p^\infty}\to 0$ $(\star_3)$
$(\star_2)$'s construction is done by horseshoe lemma. One note that $(\star_2)$'s inclusion are given by canonical inclusion $\oplus_i(p^i\Z)\to\oplus_i \Z$ whereas $(\star_3)$ is inclusion is given by $z\to (0,z)$. Now I can compute directly using resolution by seeing $0\to \operatorname{Hom}(\Z,\Z)\xrightarrow{\delta} \operatorname{Ext}^1(\Z[1/p],\Z)\to \operatorname{Ext}^1(\Z_{p^\infty},\Z)\to 0$ where $\delta$ is concretely given by inclusion pullback and this identifies $Z\to \operatorname{Ext}^1(\Z_{p^\infty},\Z)$ by $1\in \Z\to 1\in \hat{\Z}_p$. Thus I can conclude $\operatorname{Ext}^1_\Z(\Z[1/p],\Z)\cong\hat{\Z}_p/\Z\cong \Z_{p^\infty}$ where $\operatorname{Ext}^1(\Z_{p^\infty},\Z)=\hat{\Z}_p$ by pulling direct limit out.
$\textbf{Q:}$ Is it possible to identify $\operatorname{Ext}^1_\Z(\Z[1/p],\Z)\cong \Z_{p^\infty}$ without identifying the coboundary operator action? (i.e. There are too many possible ways that $\Z$ sits inside $\hat{\Z}_p$. I need to identify a particular one and quotient that out.)
I think you guys might be over-complicating things, and what Weibel had in mind is the following. Consider the short exact sequence \begin{equation*} 0 \to \mathbb{Z} \to \mathbb{Z}[1/p] \to \mathbb{Z}_{p^{\infty}} \to 0. \end{equation*} As $(\mathrm{Ext}^n(\bullet, \mathbb{Z}))_{n \in \mathbb{N}}$ is a universal cohomological $\delta$-functor, we have the following long exact sequence \begin{align*} 0 \to &\mathrm{Hom}(\mathbb{Z}_{p^{\infty}}, \mathbb{Z}) \to \mathrm{Hom}(\mathbb{Z}[1/p], \mathbb{Z}) \to \mathrm{Hom}(\mathbb{Z}, \mathbb{Z}) \to \\ &\mathrm{Ext}^1(\mathbb{Z}_{p^{\infty}}, \mathbb{Z}) \to \mathrm{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \to \mathrm{Ext}^1(\mathbb{Z}, \mathbb{Z}) \to \dots. \end{align*}
However, $\mathrm{Hom}(\mathbb{Z}[1/p], \mathbb{Z}) = 0$ because if $f \in \mathrm{Hom}(\mathbb{Z}[1/p], \mathbb{Z})$, $a \in \mathbb{Z}$ and $n \in \mathbb{N}$, then for $m \in \mathbb{N}$, $f(a/p^n) = p^mf(a/p^{m + n})$, which implies that $p^m$ divides $f(a/p^n)$ for $m \in \mathbb{N}$, which means $f(a/p^n) = 0$ and consequently, $f = 0$.
We know from the book that $\mathrm{Hom}(\mathbb{Z}, \mathbb{Z}) = \mathbb{Z}$, $\mathrm{Ext}^1(\mathbb{Z}, \mathbb{Z}) = 0$ and $\mathrm{Ext}^1(\mathbb{Z}_{p^{\infty}}, \mathbb{Z}) = \hat{\mathbb{Z}}_p$.
Thus, we have obtained the following short exact sequence \begin{equation*} 0 \to \mathbb{Z} \to \hat{\mathbb{Z}}_p \to \mathrm{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \to 0, \end{equation*} and this means $\mathrm{Ext}^1(\mathbb{Z}[1/p], \mathbb{Z}) \cong \hat{\mathbb{Z}}_p/\mathbb{Z}$ (here, we need to know how $\mathbb{Z}$ is embedded into $\hat{\mathbb{Z}}_{p^{\infty}}$, but I am sure it is in the natural way).