To take user35593's comment to fruition:
There is a unique continuous extension of the polar decomposition from $GL^+_n$ to the portion of its boundary consisting of matrices with nullity 1.
Why is the polar decomposition non-unique when $A$ is non-invertible?
Suppose $A = OP$ and $P$ has non-trivial kernel. Then if $\Omega$ is any orthogonal matrix that acts as the identity on $\mathrm{ker}(P)^\perp$, then $O\Omega P$ is another polar decomposition. And it is easy to see that this freedom is the extent of non-uniqueness: if $A = OP = \tilde{O} P$, then $O^{-1} \tilde{O}$ is an orthogonal matrix that acts as the identity on $\mathrm{ker}(P)^\perp$.
When the kernel of $A$ (equivalently kernel of $P$) is one dimensional, then the only possibility for $\Omega$ is the matrix that is the identity on $\mathrm{ker}(P)^\perp$ and $-1$ on $\mathrm{ker}(P)$.
Proof of claim
Now, since $\det A = \det O \det P$ and we know that $\det P \geq 0$, when we restrict to $GL_n^+$ we must have that $\det O > 0$. That is, on $M_n^+$ we can require the matrix $O$ in the polar decomposition to be not only orthogonal, but in fact in $SO_n$. And for rank $n-1$ matrices only one of $O$ and $O\Omega$ can fulfill that criterion. And it is easy to check that this is the continuous extension from $GL_n^+$.
Remark: if you extend from $GL_n^-$ you will pick up the other one. An illustrative example is the rank $1-1 = 0$ matrix $(0)\in M_1$. Approaching from $M_1^+$ the continuous limit of the polar decomposition gives $$ (0) = (1)(0)$$ while approaching from $M_1^-$ the continuous limit of the polar decomposition gives $$ (0) = (-1)(0). $$
Differential geometry
To answer this question in the comments: no, when $n > 1$ the set $M_n^+$ is should not be thought of as a (smooth) submanifold with boundary; it is better described as a manifold with corners. This is precisely what you outlined in your post with the different limiting directions approaching the zero matrix.
But in a neighborhood of (topological) boundary points which have rank $n-1$, the set $M_n^+$ does look like a manifold with boundary: in a neighborhood of a rank $n-1$ matrix, the function $\det: M_n\to \mathbb{R}$ has non-vanishing derivative, and so can be used as a defining function for a hypersurface.
