What happens to Km when enzyme concentration is *very* high?
Michaelis-Menten is often challenging for students because of the importance of the conditions that must exist in order for the model to hold. Walking through the derivation and the relevance of the assumptions can be helpful. Berg's Biochemistry is available via NCBI, and the section on Michaelis-Menten kinetics does a good job here, illustrating the following points:
Assumption 1:  The rates of all the relevant reactions are linear.
This assumption is what allows us to write down the equations the derivation starts with. We look at each separate reaction, and say that the rate of that reaction is equal to some rate constant times the concentration of each thing on the left side. From the equation $E + S \rightleftharpoons ES \rightarrow E + P$, we get the following

$E + S \rightarrow ES$

$Rate = k_1[E][S]$


$ES \rightarrow E + S$

$Rate = k_{-1}[ES]$


$ES \rightarrow E + P$

$Rate = k_2[ES]$



This assumption (the rate is linear) only matches the experimental evidence early on in the reaction, while most of the substrate is free $[S]$ (rather than bound up in $ES$).  Remember this. It will come into play when we evaluate assumption 3.
Assumption 2: The $ES \rightarrow E + P$ reaction is irreversible
In order for us to combine the above rate equations to derive the Michaelis-Menten equation, the only source of ES should be from $E + S \rightarrow ES$. We can't have any measurable $E + P \rightarrow ES$. Experimentally, for enzymes that follow Michaelis Menten kinetics, this is true when $[E]$ and $[P]$ are small relative to [ES]. So here we have our first requirement that $[E]$ be small. Biochemical reactions tend to not involve very high changes in free energy. The in situ free energy of ATP hydrolysis is the high end for a reaction in the cell, and that releases about $50kJ mol^{-1}$. We're not talking about combustion here. So in fact, these reactions are often reversible at appreciable levels of $[E]$ and $[P]$
Assumption 3: ES formation and breakdown is at steady state
This assumption is often described in a confusing way, i.e. the entire reaction ($E + S \rightarrow E + P$) is at steady state. Lets drill down to what we're actually talking about. In order to derive the MM equation. We need the rate of formation of $ES$ to equal the rate of breakdown of $ES$, that is, for $\frac{d[ES]}{dt} = 0$. That is steady state. Let's say that in words: the change in $[ES]$ needs to be zero. We only achieve steady state at an early point in the reaction (remember, we need most of the substrate to be free $[S]$ for assumption 1 to hold), if $[S]$ is substantially greater than $[E]$. $E$ needs to be saturated with $S$ while $S$ is still mostly in the free form. This is the key combination of assumptions that the situation you described violates.
Once we meet the three assumptions above, we can write down the following equality:
$k_1[E][S] = (k_{-1} + k_2)[ES]$
This equation says that the rate of formation of $ES$ (the rate from equation 1 at the top) is equal to the combined rate of the breakdown of $ES$, either into $E + S$ or $E + P$ (equations 2 and 3 at the top). Once we have this equality, we can derive the Michaelis-Menten equation: $V_o = V_{max} \frac{[S]}{[S] + K_m}$, where $K_m = \frac{k_{-1} + k_2}{k_1}$. Again, see Biochemistry to walk through this.