The following provides a counterexample.
$\newcommand\R{\mathbb{R}}\newcommand\norm[1]{\lVert#1\rVert}\newcommand\abs[1]{\lvert#1\rvert}\newcommand\d{\mathrm{d}}\newcommand\MM{\mathcal{M}}\newcommand{\floor}[1]{\lfloor #1 \rfloor}\newcommand{\dual}[2]{\langle #1 ,\, #2 \rangle}$Let $\Omega = B_1(0) \subset \R^2$ be the (open) unit ball.
We denote by $\delta_{\hat r}$ the line measure
which is located at the radius $\hat r \in (0,1)$ and with mass $2 \, \pi$ (i.e., line density $1/\hat r$).
By $(-\Delta_0)^{-1}$ we denote the solution mapping associated with the Laplace equation with homogeneous Dirichlet boundary condition on $\Omega$.
We have
\begin{equation*}
v_{\hat r}(x,y)
:=
(-\Delta_0)^{-1}(\delta_{\hat r}) (x,y)
=
\begin{cases}
\log(1 / \hat r), & \text{if } r \le \hat r , \\
\log(1 / r), & \text{if } r > \hat r .
\end{cases}
\end{equation*}
Here and in the sequel, we use $r = \sqrt{x^2 + y^2}$.
We find
\begin{equation*}
\frac{\partial}{\partial r} v_{\hat r}(x,y)
=
\begin{cases}
0 & \text{if } r \le \hat r \\
-1/r & \text{if } r > \hat r
\end{cases}
\end{equation*}
and, thus,
\begin{equation*}
\norm{v_{\hat r}}_{H_0^1(\Omega)}^2
=
\int_\Omega \abs{\nabla v}^2 \, \d (x,y)
=
\int_\Omega (\frac{\partial}{\partial r} v)^2 \, \d (x,y)
=
\int_{\hat r}^1 1/r \, \d r
=
\log(1/\hat r).
\end{equation*}
Now, let $q \in (0,1)$ and a sequence $\{c_i\}_{i = 1}^\infty \subset \R^+$ be given.
We set $r_i = q^i$.
We define a sequence $\{\mu_k\} \subset H^{-1}(\Omega) \cap \MM(\Omega) = (H_0^1(\Omega))^* \cap (C_0(\Omega))^*$
by
\begin{equation*}
\mu_k
:=
\sum_{i = 1}^k
c_i \, (\delta_{r_{2 \, i}} - \delta_{r_{2\,i-1}})
.
\end{equation*}
Since all line measures have mass $2 \, \pi$,
the sequence $\{\mu_k\}$ is Cauchy in $\MM(\Omega)$ iff $\{c_i\}$ is summable.
In order to compute the $H^{-1}(\Omega)$-norm of $\mu_k$, we set
\begin{equation*}
v_k := (-\Delta_0)^{-1} \mu_k
\end{equation*}
and have
\begin{equation*}
\norm{v_k}_{H_0^1(\Omega)} = \norm{\mu_k}_{H^{-1}(\Omega)}.
\end{equation*}
Since
\begin{equation*}
\frac{\partial}{\partial r} v_k(x,y)
=
\begin{cases}
-c_i/r & \text{if } r_{2\,i} \le r \le r_{2\,i-1} \text{ with } i \in \{1,\ldots,k\},
\\
0 & \text{else},
\end{cases}
\end{equation*}
we find
for $n \le k$
\begin{align*}
\norm{\mu_n - \mu_k}_{H^{-1}(\Omega)}^2
&=
\norm{v_n - v_k}_{H_0^1(\Omega)}^2
\\
&=
\int_\Omega \abs{\nabla (v_n - v_k)}^2 \, \d (x,y)
\\
&=
\int_\Omega (\frac{\partial}{\partial r} (v_n - v_k))^2 \, \d (x,y)
\\
&=
\sum_{i = n + 1}^k c_i^2 \, \int_{r_{2\,i}}^{r_{2\,i-1}} 1/r \, \d r
\\
&=
\sum_{i = n + 1}^k c_i^2 \, \log(r_{2\,i-1}/r_{2\,i})
\\
&=
\log(1/q) \,
\sum_{i = n + 1}^k c_i^2
\end{align*}
Hence, the sequence $\{\mu_k\}$ is Cauchy in $H^{-1}(\Omega)$
iff $\{c_i\}$ is square-summable.
In case $\{c_i\}$ is summable,
the limits of $\mu_k$ in $H^{-1}(\Omega) = H_0^1(\Omega)^*$ and $\MM(\Omega) = C_0(\Omega)^*$
coincide, since $C_0^\infty(\Omega)$ is a dense subspace of $H_0^1(\Omega)$ and of $C_0(\Omega)$.
Now,
we choose $c_i = i^p$ for some $-3/2 < p < -1$.
Then $c_i$ is summable and square summable.
The positive part of
\begin{equation*}
\mu
=
\sum_{i = 1}^\infty c_i \, (\delta_{r_{2\,i}} - \delta_{r_{2\,i-1}})
\end{equation*}
is
\begin{equation*}
\mu^+
=
\sum_{i = 1}^\infty c_i \, \delta_{r_{2\,i}}.
\end{equation*}
Note that $\mu$ belongs to $\MM(\Omega)$ and $H^{-1}(\Omega)$.
Let $\varphi \in C_0^\infty(\Omega)$ with $0 \le \varphi \le 1$
and $\varphi \equiv 1$ on $B_q(0)$ be given.
For $0 < s < 1/2$,
the function
\begin{equation*}
v(x,y)
=
\log
\big(1/r\big)^s
\,
\varphi(x,y)
\end{equation*}
belongs to $H_0^1(\Omega)$.
For given $K > \log(1/q)^s$, we consider the truncation $v_K$ of $v$ at $K$
and have
\begin{equation*}
v_K(x,y)
=
\begin{cases}
K & r \le \exp(-K^{1/s}), \\
v(x,y) & r > \exp(-K^{1/s}).
\end{cases}
\end{equation*}
Moreover, $v_K \to v$ in $H_0^1(\Omega)$ as $K \to \infty$.
But
\begin{equation*}
\dual{\mu^+}{v_K}
=
2\,\pi\,\sum_{i = 1}^\infty c_i \, v_K(r_i)
\ge
2\,\pi\,\sum_{i = 1}^{n(K)}
c_i \,
\log
\big(1/q^i\big)^s
=
2\,\pi\,
\log
\big(1/q\big)^s
\,
\sum_{i = 1}^{n(K)} i^{p + s}
,
\end{equation*}
where $n(K) = \floor{K^{1/s} / \log(1/q)}$.
Note that $n(K) \to \infty$ as $K \to \infty$ and, hence,
\begin{equation*}
\dual{\mu^+}{v_K}
\ge
2\,\pi\,
\log
\big(1/q\big)^s
\,
\sum_{i = 1}^{n(K)} i^{p + s}
\to
\infty
\end{equation*}
as $K \to \infty$ if $p + s \ge -1$.
Note that for all $p \in (-3/2,-1)$, we can choose $s \in (0,1/2)$
such that $p + s \ge -1$.
This shows that
$\mu^+ \in \MM(\Omega)$
is not bounded on
$H_0^1(\Omega) \cap C_0(\Omega)$
w.r.t.\ the $H_0^1(\Omega)$-norm.
I would be glad if anyone who checked the counterexample would leave a comment.
