The notation $\mathbf Z(p^\infty)$ and the name Prufer group are standard for some people, but I can never remember what these things mean. So let's write this group in another way.
The group $\mathbf Z(p^\infty)$ is isomorphic to the group of $p$-power roots of unity. The group of all roots of unity is isomorphic to $\mathbf Q/\mathbf Z$: in $\mathbf C$ we can write roots of unity as $e^{2\pi ir}$ where $r$ is rational and $r$ is determined from $e^{2\pi ir}$ up to addition by an integer. The $p$-power roots of unity are the elements of $p$-power order in the roots of unity, so $\mathbf Z(p^\infty)$ can be regarded as the elements of $p$-power order in $\mathbf Q/\mathbf Z$, which is all $r \bmod \mathbf Z$ where $r$ has a $p$-power denominator. Rationals with $p$-power denomiantor form the ring $\mathbf Z[1/p]$, so
$$
\mathbf Z(p^\infty) \cong \mathbf Z[1/p]/\mathbf Z.
$$
Representatives of $\mathbf Z[1/p]/\mathbf Z$ are finite sums $c_{-n}/p^n + \cdots + c_1/p$ where $0 \leq c_i \leq p-1$, which behave under addition mod $\mathbf Z$ like $p$-adic numbers with the $p$-adic integer part ignored. Thus
$$
\mathbf Z(p^\infty) \cong \mathbf Z[1/p]/\mathbf Z \cong \mathbf Q_p/\mathbf Z_p.
$$
So from now on I am going to ignore the names and notation you used, and instead show
$$
\mathbf Q_p \cong \varprojlim \mathbf Q_p/\mathbf Z_p
$$
where the maps $\mathbf Q_p/\mathbf Z_p \to \mathbf Q_p/\mathbf Z_p$ are multiplication by $p$.
When $n$ is a positive integer, we have the commutative diagram
$$
\require{AMScd}
\begin{CD}
\mathbf Q_p/\mathbf Z_p @>{x \to px}>> \mathbf Q_p/\mathbf Z_p\\
@VVV @VVV\\
\mathbf Q_p/p^{n+1}\mathbf Z_p @>{{\rm redn.}}>> \mathbf Q_p/p^n\mathbf Z_p
\end{CD}
$$
where the top map is the transition map in the inverse limit you are asking about, the bottom map is the reduction map $x \bmod p^{n+1}\mathbf Z_p \mapsto x \bmod p^n\mathbf Z_p$ (which makes sense since $p^{n+1}\mathbf Z_p \subset p^n\mathbf Z_p$), and the vertical maps are group isomorphisms where you multiply by the obvious power of $p$. Check the diagram commutes by starting with an $x \bmod \mathbf Z_p$ in the upper left and following the diagram around both ways to the lower right, each way getting $p^{n+1}x \bmod p^n\mathbf Z_p$.
Because the vertical maps are group isomorphisms, the inverse limits along the top and bottom over all $n \geq 0$ are isomorphic groups. You asked about the inverse limit along the top, but let's work instead with the inverse limit along the bottom. Writing an inverse limit in the right way can make it much clearer what is going on. A nice example of this idea in another MSE question is here.
Unlike with the setup you were looking at, I think it is intuitively plausible that the inverse limit along the bottom should be $\mathbf Q_p$ as a group because as $n$ grows, the groups $p^n\mathbf Z_p$ shrink to $0$. The process of lifting your way through the inverse limit along the bottom of the commutative diagram amounts to revealing more and more of the $p$-adic expansion of a $p$-adic number $x$: think about what you can see in $x \bmod p^3 \bmod \mathbf Z_p$ compared to $x \bmod p^{10}\mathbf Z_p$.
An element in the bottom inverse limit is a sequence $\{x_n \bmod p^n\mathbf Z_p\}$ where $x_n \in \mathbf Q_p$ for all $n \geq 0$ and
$x_{n+1} \equiv x_n \bmod p^n\mathbf Z_p$ for all $n$. That congruence says $|x_{n+1} - x_n|_p \leq 1/p^n$, so $x_{n+1} - x_n \to 0$. Thanks to the non-archimedean nature of the $p$-adic absolute value, a sequence whose consecutive differences tend to $0$ is Cauchy. Thus $\{x_n\}$ is a Cauchy sequence in $\mathbf Q_p$, so it has a limit in $\mathbf Q_p$ by completeness of $\mathbf Q_p$. If we use a different set of representatives for the same element of the inverse limit, say $\{y_n \bmod p^n\mathbf Z_p\}$ where $x_n \equiv y_n \bmod p^n\mathbf Z_p$ for all $n$, check that the limit of the $y_n$'s in $\mathbf Q_p$ is the limit of the $x_n$'s.
Thus we get a well-defined mapping $f \colon \varprojlim \mathbf Q_p/p^n\mathbf Z_p \to \mathbf Q_p$ by $f(\{x_n \bmod p^n\mathbf Z_p\}) = \lim_{n \to \infty} x_n$. Check $f$ is a group homomorphism. Setting $x =
\lim_{n \to \infty} x_n$, we have $x \equiv x_n \bmod p^n\mathbf Z_p$ for all $n$ since the inverse limit consistency condition $x_{n+1} \equiv x_n \bmod p^n\mathbf Z_p$ for $n \geq 0$ implies $x_m \equiv x_n \bmod p^n\mathbf Z_p$ for all $m \geq n$, so we have $x \equiv x_n \bmod p^n\mathbf Z_p$ for all $n$ by letting $m \to \infty$.
There is an obvious map $g \colon \mathbf Q_p \to \varprojlim \mathbf Q_p/p^n\mathbf Z_p$, namely $g(x) = \{x \bmod p^n\mathbf Z_p\}$. Easily $f(g(x)) = x$ for all $x \in \mathbf Q_p$ and the previous paragraph shows $g(f(\{x_n \bmod p^n\mathbf Z_p\})) = \{x_n \bmod p^n\mathbf Z_p\}$. Thus $f$ and $g$ are inverses of each other, so the inverse limit of $\mathbf Q_p/p^n\mathbf Z_p$ with the natural reduction maps is isomorphic as a group to $\mathbf Q_p$.
Best Answer
You have shown that $-\frac{1}{10} = \ldots\!111_{11}$. Which is true, no problems here! It sounds like you are worried that $(-1)\times \ldots\!111_{11} = \ldots\!(10)(10)(10)_{11} \times \ldots\!111_{11}$ gives $-1$? Well, it doesn't... maybe you are multiplying pointwise (term by term) by mistake?
Instead of working out the result of $\ldots\!(10)(10)(10)_{11} \times \ldots\!111_{11}$, it's easier to do something like: $$ -\frac{9}{10} = 9\cdot \left(-\frac{1}{10}\right) = \ldots\!999_{11} $$ $$ \implies \frac{1}{10} = -\frac{9}{10} + 1 = \ldots\!999(10)_{11} $$ It should also be possible to verify directly that $\ldots\!(10)(10)(10)_{11} \times \ldots\!111_{11} = \ldots\!999(10)_{11}$.