$f:D \to \mathbb R$ is an injective function on $D \subset \mathbb R$, so the inverse function $f^{-1}$ exists.
$a \in \mathbb R$ is such that in any neighbourhood of $a$ there is a point form $D$, so we can talk about $\lim_{x \to a}f(x)$ and we assume that $\lim_{x \to a}f(x)=b$.
In any neighbourhood of $b$ there is a point from $D':=f(D)$.
Indeed, $\lim_{x \to a}f(x)=b \iff \forall \epsilon >0, \exists \delta > 0: f(D \cap \mathcal U_\delta(a)) \subset \mathcal U_\epsilon(b)$. We take $x_0 \in D \cap \mathcal U_\delta(a)$ and see that $f(x_0) \in D' \cap \mathcal U_\epsilon(b)$. Because $\epsilon$ was freely chosen we see that in any neighbourhood of $b$ there is a point from $D':=f(D)$.
So the observation above makes it possible to analyze $\lim_{y \to b}f^{-1}(y)$.
What minimal assumptions w.r.t. $f$ and $D$ should I make (continuity, monotonicity, interval) so that $\lim_{y \to b}f^{-1}(y)=a$.
First of all you are implicitly assuming that the limit at $a$ exists; in other words, $f$ should have a continuous extension to $D\cup\{a\}$ and if $a\in D$ then $f$ should be continuous at $a.$ If you want conditions valid for all $a\in\overline D$ then you must assume that $f$ is continuous (and has limits on the boundary of its domain) before you can even formulate the problem. You also want that the continuous extension $\overline f$ of $f$ to $\overline D$ is still one-to-one.
You are effectively asking for conditions under which the inverse of $\overline f$ is continuous, as well; in other words: $\overline f$ should be a homeomorphism between $\overline D$ and $\overline{f(D)}.$
A sufficient additional condition is then that $\overline D$ be compact: continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms.
As a counterexample for noncompact closed domain let $D=\mathbb N,$ $f(0)=0,$ $f(n)=\frac1n$ ($n>0$). Then the inverse $f^{-1}$ is discontinuous at $0.$