I think I can give a partial answer.

*(I will use the OP's notation of + for the adjoint)*

1) It is clear that any "good" matrix has every diagonal entry equal to 1.

2) If A is "good", then it is positive. Indeed, being good means that there exist $k\in\mathbb{N}$ and $U_1,\ldots,U_n$ unitaries in $M_k(\mathbb{C})$ with
```
\[
A=\frac1k\,\mbox{tr}^{(n)}\left(\begin{bmatrix}U_1\\ U_2\\
\vdots \\ U_n\end{bmatrix} \,
\begin{bmatrix}U_1\\ U_2\\
\vdots \\ U_n\end{bmatrix}^+\right),
\]
```

where $\mbox{tr}^{(n)}$ is the map that replaces each $k\times k$ block by its trace. Since
the trace has abelian range it is completely positive, so $\mbox{tr}^{(n)}$ is positive and thus
$A$ is positive.

3) For $n=2$, the converse holds (i.e. any positive matrix with 1 in the diagonal is "good"). Indeed, let $A=\begin{bmatrix}1&a\\ a&1\end{bmatrix}$, with $|a|\leq1$. Let $k=2$, $U_1=I_2$,
$U_2=\begin{bmatrix}a&\sqrt{1-|a|^2}\\ -\sqrt{1-|a|^2}&a\end{bmatrix}$. Then
```
\[
A=\frac12\,\mbox{tr}^{(n)}\left(\begin{bmatrix}U_1\\ U_2\end{bmatrix} \,
\begin{bmatrix}U_1\\ U_2\end{bmatrix}^+\right).
\]
```

4) So the question remains, is every $n\times n$ positive matrix with diagonal 1, "good"?
For $n\geq3$, I still cannot see too much in this direction. On the one hand, being free to choose $k$ gives an incredible amount of unitaries to play with. On the other, positivity is hard to make precise as a relation among the coefficients, and so I cannot see an obvious way to choose the unitaries in a satisfactory way (somehow the restriction given by positivity should appear in the choice of the unitaries, and I couldn't make sense of it).

5) If the conjecture were true (and it is for $n=2$) this makes the set $S$ closed (a norm-limit of positive matrices with diagonal one is going to be again positive with diagonal 1). Otherwise, from the definition alone, I don't really see it.