Partial path known in Single source shortest path problem
Your question is strongly related to the idea of macro-moves which are defined as follows: $M(p, q)=\langle p, n_1, n_2, \ldots, n_k, q\rangle$, i.e., as a list of successive states which are known to get to $q$ from $p$. Additionally, macro moves might guarantee that the path they store is also optimal, i.e., that the path $\langle p, n_1, n_2, \ldots, n_k, q\rangle$ is the shortest path between $p$ and $q$. Let us denote such macro-moves as $M^*(p, q)$.

In the following I'll assume you are interested in computing the optimal path from a start state $s$ to an arbitrary goal state $t$ ---as you actually mention the "shortest path" at the beginning.

So now, regarding your question:


In case that $M^*(p,q)$ is known to be a subpath of the optimal path from $s$ to $t$, or if alternatively, it is known that the optimal path from $s$ to $t$ does not go through $p$, then you can safely substitute each generation of $p$ in OPEN by $q$ with a cost equal to the cost of the path to generate $p$ plus the cost of the path $M^*(p,q)$.
In case none of the preceding conditions hold then you can still use the macro-moves but you have then to store each node in $M(p,q)$, but $p$ of course, into the OPEN list ---i.e., the intermediate milestones you mention. 


Hold on! You might think then that the second case is a waste of time/space, but it ain't if you break ties in favour of nodes with larger $g$-values, i.e., while nodes in OPEN are sorted in increasing order of their $f$-value, breaking ties in favour of nodes with larger $g$-values place those nodes with larger $g$-values first ---within the same $f$-layer. This way, in case that an optimal path (as there might be several paths with the same cost and equal to the minimum cost) goes through $p$ there is an opportunity that you will find it without generating other descendants.

To see why, consider that a subpath of $M^*(p,q)$ is part of the optimal path from $s$ to $t$, $\langle p, n_1, n_2, ...,n_\ell\rangle$, such that $\ell\leq k$, and let $g(n)$ denote the cost of the path from $s$ to $n$ and $g(p, n_i)$ denote the cost of the path $\langle p, n_1, n_2, ..., n_i\rangle$ where $\langle p, n_1, n_2, ..., n_i\rangle\subset M^*(p,q)$. The second case suggests to insert into open $n_1, n_2, ..., n_\ell$ each with a cost equal to $g(p)+g(p,n_i)$, $1\leq i\leq\ell$. If the heuristic function is consistent (as you mention) there is a good opportunity that $f(n_\ell)=f(n_i)$, $\forall i, 1\leq i&lt;\ell$. Now, breaking ties in favour of nodes with larger $g$-values, $n_\ell$ will be placed before the others so that A$^*$ will expand it first. If $f(n_\ell)$ equals the cost of the optimal path then A$^*$ will find the optimal solution with out expanding nodes $n_i$.

Maybe you think there are many "maybes and ifs" here but consider: first, breaking ties in favour of larger $g$-values is customary practice; second, inserting the intermediate states of $M^*(p, q)$ causes no harm so that proceeding as in the second case is usually a good idea.

Hope this helps,