There is probably not an isomorphism.

First, note a reason to be suspicious - if you had such an isomorphism, couldn't you just iterate it to get an isomorphism $\overline{M}_{g,A}=\overline{\mathcal M}_{g,n}$. This would make the work of Brendan Hasset in his paper somewhat superfluous.

I claim there is a natural map $ \overline{M}_{g, A \cup \{a_n+1 \} } \to \mathcal C_{g,A} $. To construct it, simply delete the last marked point. If the twisted canonical bundle ceases to be ample on the component the last marked point was on, contract it. Repeat until you have a weighted stable curve with $n$ marked points. Then choose from among the points of this curve the image of the last marked point under the contraction map.

However, this map often fails to be injective. Suppose there is some $a_i, a_j$ with $a_i+a_j\leq 1$, $a_i + a_j + a_{n+1} >1$. Consider a genus $g$ smooth curve glued to a single copy of $\mathbb P^1$ where points $i,j,n+1$ are on the $\mathbb P^1$ and the other points are distributed on the genus $g$ curve. Removing point $a_{n+1}$ causes the $\mathbb P^1$ to be contracted, and $a_i$ and $a_j$ become the same point. This happens regardless of the location of $a_i, a_j, a_{n+1}$ on the old $\mathbb P^1$. But there is a 1-parameter family of possible locations of these three points and the one node up to isomorphism. This curve is contracted to a point by the map $ \overline{M}_{g, A \cup \{a_n+1 \} } \to \mathcal C_{g,A} $.

Because the other points can be chosen on the genus $g$ curve to kill all the automorphisms, passing to the coarse moduli space does not change anything in this example.