An equilibrium reaction in the presence of other fast reactions -- is it necessary also to consider the individual rate constants?
You're correct. If other reactions are occurring that are fast on the time scale of the equilibrium reactions, then you must treat the forward and reverse reactions separately, including by using their separate rate constants.

"Equilibrium" is a mathematical model that never actually perfectly matches reality.  In order for most systems to actually be at "perfect equilibrium" we would have to let them sit and react for an infinite amount of time. Of course, practically speaking, that's not necessary: equilibrium is reached to within measurement precision in quite reasonable amounts of time for many reactions. What's necessary in order for the concept of 'equilibrium' to be useful in analyzing a reaction system is for the time scale of those reactions to be "very short" (or, equivalently, for the rate of those reactions to be "very fast") when compared to all other reactions of interest in the system.

In other words, when we write the following system of reactions:

$$
\ce{A + B &lt;=&gt;[K_1] C + D} \\ ~ \\
\ce{C + E -&gt;[k_2] P}
$$

what we're really writing is:

$$
\ce{A + B \underset{k_{-1}}{\overset{k_1}{\rightleftarrows}} C + D} \\ ~ \\
\ce{C + E -&gt;[k_2] P} \\ ~ \\
\begin{align}
k_1\ce{[A][B]} &amp;\gg k_2\ce{[C][E]} \\
k_{-1}\ce{[C][D]} &amp;\gg k_2\ce{[C][E]} \\
K_1 &amp;= {k_1 \over k_{-1}}
\end{align}
$$

The relative values of the forward and reverse rate constants can vary, with their ratio described by $K_1$, but at all points through the reacting process of interest, both the forward and reverse equilibrium reactions are assumed to be much faster than the final, product-forming reaction. If this assumption doesn't hold, then we can't treat the first reaction as "just" an equilibrium reaction.