Here is a positive answer, for the interior points of the set $M_k:=\mathcal M_k$. Let indeed $c=(c_1,\dots,c_k)$ be any point in the interior of $M_k$.
Let $P$ stand for the set of all probability measures on $[0,1]$.
Let us show that then there is an absolutely continuous measure $\nu\in P$ such that $m_j(\nu)=c_j$ for $j=1,\dots,k$.

For each $\mu\in P$ and each real $h\in(0,1/2)$, define $\mu_h\in P$ by the convolution-like condition that
$$\int_{[0,1]} f\,d\mu_h=\int_{[0,1/2]}\mu(dx)\int_0^h\frac{du}h\,f(x+u)
+\int_{(1/2,1]}\mu(dx)\int_0^h\frac{du}h\,f(x-u) \tag{1}
$$
for all nonnegative Borel functions $f$ on $[0,1]$. Then $\mu_h$ is clearly absolutely continuous, with density
$$[0,1]\ni y\mapsto\frac1h\,\mu\big((y-h,y)\cap[0,1/2]\big)+\frac1h\,\mu\big((y,y+h)\cap(1/2,1]\big).$$
Since $\mu\mapsto\mu_h$ is a linear operator, the set
$$M_{k,h}:=\{(m_1(\mu_h),\dots,m_1(\mu_h))\colon\mu\in P\}
$$
is convex.

It is enough to show that $c=(c_1,\dots,c_k)\in M_{k,h}$ for small enough $h$. Suppose the contrary. Then there is a hyperplane $H$ in $\mathbb R^k$ through the point $c$ such that the set $M_{k,h}$ is contained in a closed half-space $H_+$ whose boundary is $H$. Since $c$ is in the interior of $M_k$, there is a sphere $S_c(r)$ of positive radius $r$ centered at $c$ such that $S_c(r)\subseteq M_k$. Take the point $b=(b_1,\dots,b_k)\in S_c(r)$ that is at distance $r$ from $H_+$ and hence at distance $\ge r$ from the set $M_{k,h}$.
Since $b\in S_c(r)\subseteq M_k$, there is some measure $\mu\in P$ such that $m_j(\mu)=b_j$ for all $j=1,\dots,k$.

Since $|(x+u)^j-x^j|\le j|u|\le k|u|$ for $j=1,\dots,k$ and $x,x+u$ in $[0,1]$, by $(1)$ we have $|m_j(\mu_h)-b_k|=|m_j(\mu_h)-m_j(\mu)|\le kh/2$ for $j=1,\dots,k$, so that the distance from the point $(m_1(\mu_h),\dots,m_k(\mu_h))$ to $b$ is $\le k^{3/2}h/2<r$ if $h<2r/k^{3/2}$, which contradicts the condition that $b$ is at distance $\ge r$ from the set $M_{k,h}$. QED

**Addendum:** For a point $c=(c_1,\dots,c_k)$ to be in the interior of $M_k$, it is enough that $c_j=m_j(\mu)$ for some measure $\mu\in P$ with at least $k+1$ points of support and all $j=1,\dots,k$; cf. e.g. the lemma in my answer to the question at **[moment-matching]**.
This condition of $\mu$ having at least $k+1$ points of support is however not necessary; indeed, for large $k$, the measure $\mu$ representing the point $c$ may have only about $k/2$ points of support -- but those points need to be specially chosen; namely, the points need then to be the so-called roots of a principal representation of $c$; see e.g. Definitions 3.1 and 3.2 and Corollary 3.1 in Ch. II in **[Karlin--Studden]**.

**Addendum 2:** In particular, it follows from the first sentence of the above Addendum that the condition that the point $c$ lie in the interior of $M_k$ is, not only sufficient, but also necessary for $c$ to be representable by an absolutely continuous probability measure. That is, $c$ is in the interior of $M_k$ if and only if $c_j=m_j(\mu)$ for some absolutely continuous $\mu\in P$ and all $j=1,\dots,k$.