Can classical Lagrangian mechanics be obtained directly from energy conservation? Is there a way to derive classical Lagrangian mechanics (in particular, the classical Lagrangian $L = T-V$ and the Euler-Lagrange equation), under the simple assumption that mechanical energy is conserved?
Edit: Since it seems that my question caused some confusion, I will attempt to clarify.
Assume that we know the work-energy theorem (and thereby, know that the sum of the kinetic and potential energies, $T+V$, is conserved). Using this knowledge, is there a direct, intuitive way to derive classical Lagrangian mechanics (which amounts to deriving the Euler-Lagrange equation)?
 A: 
Let's say that the potential and kinetic energies of a given classical system at time $t$ are given by the functions $V(t)$ and $T(t)$, respectively. The energy state of the system can then be represented by a vector
\begin{equation} \tag{1} \label{E1}
 \mathbf{E}(t) \equiv V(t)\mathbf{\hat{V}} + T(t)\mathbf{\hat{T}}
\end{equation}
in a two-dimensional energy space spanned by the unit vectors $\mathbf{\hat{V}}$ and $\mathbf{\hat{T}}$ (see figure). Since energy is conserved, we have that
\begin{equation} \tag{2}
 V(t) + T(t) = H
\end{equation}
where $H$ is constant, such that $\mathbf{E}(t)$ is confined to the corresponding line in energy space. This implies that (\ref{E1}) can equally well be written as
\begin{equation} \tag{3} \label{E2}
 \mathbf{E}(t) = \frac{1}{\sqrt{2}}\left[H\mathbf{\hat{H}} + L(t)\mathbf{\hat{L}}\right]
\end{equation}
with the new basis
\begin{align}
 \mathbf{\hat{H}} & \equiv \frac{1}{\sqrt{2}}(\mathbf{\hat{T}} + \mathbf{\hat{V}}), \tag{4} \\
 \mathbf{\hat{L}} & \equiv \frac{1}{\sqrt{2}}(\mathbf{\hat{T}} - \mathbf{\hat{V}}) \tag{5}
\end{align}
and the single energy variable
\begin{equation} \tag{6}
 L(t) \equiv T(t) - V(t)
\end{equation}
which we recognize as the classical Lagrangian of the system. Here, $L(t)$ can be thought of as representing the time-dependent part of the system's energy trajectory.
Now, a fundamental assumption of classical physics is that trajectories are unique with respect to initial conditions, such that if we know the state of a system at some time $t_0$, it is possible to uniquely determine its state at any later time $t_1$. This means that the average energy displacement over such a time interval
\begin{equation} \tag{7} \label{Ebar}
 \bar{\mathbf{E}} \equiv \frac{1}{t_1-t_0}\int_{t_0}^{t_1} \mathbf{E}(t) \, \mathrm{d} t
\end{equation}
must be some specific vector, such that any arbitrarily small perturbation $\delta\bar{\mathbf{E}}$ must vanish;
\begin{equation} \tag{8} \label{dE}
 \delta\bar{\mathbf{E}} = \mathbf{0}.
\end{equation}
If we now use (\ref{E2}) and (\ref{Ebar}) in (\ref{dE}) and discard constant terms and factors, we get that
\begin{equation} \tag{9}
 \delta\int_{t_0}^{t_1} L(t) \, \mathrm{d} t = 0
\end{equation}
which is equivalent to the Principle of stationary action, from which the Euler-Lagrange equation can be obtained in the usual manner through Calculus of Variations.
A: The following is a comment, I'm posting it in answer space since it is too large for the comment space.

In a Jan. 2020 answer to a question about derivation in physics stackexchange contributor Kevin Zhou (@knzhou) pointed out the following:

[...] in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X).

I'm quoting that statement here to emphasize the following: while it is the case that from assumption of conservation of energy the logical implications that you describe do follow, no wider conclusions can be drawn. The logical steps are correct, but they do not shed new light.
In any logical system there is great freedom to interchange theorem and axiom without changing the content of the system.


The Work-Energy theorem is obtained by deriving it from F=ma
The Work-Energy theorem:
$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2  $$
It follows from the work-energy theorem that in the course of physics taking place the rate at which the kinetic energy is changing will be equal to the rate at which the potential energy is changing - with opposite sign.
Hence: the value of the sum of kinetic energy and potential energy will remain the same.

Next:
There is a close connection between the Work-Energy theorem and Hamilton's stationary action.
Both are expressed in terms of kinetic and potential energy.
From both $F=ma$ can be recovered.
In fact:
In cases where the work-energy theorem holds good: Hamilton's stationary action holds good also.


A: It seems OP's question (v2) is ill-posed in the sense that it is not completely clear what is known and what is not, and under which conditions. E.g. do we know the mechanical energy as a function $$(q,\dot{q})\quad\mapsto\quad T(q,\dot{q})+V(q)~?\tag{1}$$
If yes it seems that in many case (especially if one has a geometric understanding of the system), one can deduce the kinetic energy $T$ and potential energy $V$ independently, up to irrelevant constants. And then one would essentially have the Lagrangian $L=T-V$ from the very beginning, rendering OP's question moot.
