I cannot be the the first person to notice that $ \frac{dE}{dv} = p$ (or, alternatively, that $\int pdv = E$).
The Hamiltonian equations of motion are:
$$
\frac{\partial H}{\partial \vec p} = \frac{d\vec x}{dt}=\vec v \tag{1}
$$
and
$$
\frac{\partial H}{\partial \vec q} = -\frac{d\vec p}{dt} \tag{2}\;,
$$
where $H$ is the energy, which usually corresponds to the total mechanical energy, which is the kinetic energy plus the potential energy.
You seem to be using the symbol $E$ for kinetic energy. I will use the symbol $T$ for kinetic energy, since $E$ is often reserved for total energy.
Why on earth would it be the case the momentum is the derivative of kinetic energy with respect to velocity?
It is not always, but often. What is true is: The velocity is the derivative of the energy with respect to the momentum:
$$
\vec v = \frac{\partial H}{\partial \vec p}\;.
$$
In the case where the momentum $\vec p$ is equal to the mass $m$ times the velocity $\vec v$ (which is not always the case, but is often the case):
$$
v_i = \frac{\partial H}{\partial p_i} = \sum_j\frac{\partial H}{\partial \vec v_j}\frac{\partial v_j}{\partial p_i} = \frac{\partial H}{\partial \vec v_i}\frac{1}{m}\;.
$$
Or, moving the $m$ to the other side, in this case where $\vec p = m\vec v$,
we see:
$$
\vec p = \frac{\partial H}{\partial \vec v}
$$
Now, assuming the potential energy $U$ doesn't depend on the velocity:
$$
H(\vec x, \vec p) = T(\vec p) + U(\vec x)\;.
$$
So, we can further write:
$$
\vec p = \frac{\partial H}{\partial \vec v}=\frac{\partial T}{\partial \vec v}
$$
Or, maybe the better question is, is there a physical meaning to the fact that the derivative of kinetic energy with respect to velocity is momentum (or that the integral of momentum over velocity is kinetic energy)?
As mentioned above, this is not an unqualified fact. It is sometimes/often true.
You may be aware of the work-kinetic-energy theorem. This says that the work done by the net force is equal to the change in kinetic energy:
$$
\delta T = \int d\vec x \cdot \vec F = \int d\vec x \cdot \frac{d\vec p}{dt} = \int \vec v\cdot d\vec p\;.
$$
If we can write $\vec p = m\vec v$ then we again arrive at your conclusion:
$$
\delta T = \int \vec p \cdot d\vec v
$$
