Let's first clarify which formulation of Noether's theorem we'll be using:
The Lagrangian will be a function $$ L = L(x,v,t) $$ and the action a functional $$ S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t),t) \,dt $$
Proposition. If the transformation $$ t\to t'(t) = t + \epsilon T(t) $$ $$ x\to x'(x,t) = x + \epsilon X(t)$$ $$ q'(t') = q(t(t')) + \epsilon X(t(t')) $$ is a quasi-symmetry of the action $$ \delta S \approx \Delta K $$ on-shell (ie assuming the equations of motion), then there is a conserved quantity $$ \frac{d}{dt} \left( \frac{\partial L}{\partial v} (X - \dot q T) + LT - K \right) \approx 0 $$ Here, $$ \delta S = \frac{d}{d\epsilon}\Big|_{\epsilon=0} S[q'] $$ $$ \Delta K = K(t_2)-K(t_1) = \int_{t_1}^{t_2}\frac{dK}{dt} dt $$
Proof. Given in this answer.
We also need a result from the body of the proof, namely that $$ \delta S = \int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial x}X + \frac{\partial L}{\partial v} (\dot X - \dot q \dot T) + \frac{\partial L}{\partial t} T + L \dot T\right]\,dt $$
Now, there are two ways to arrive at $$ \frac{d}{dt} \left( \frac{\partial L}{\partial v} \dot q - L \right) \approx 0 $$
First, we can choose $X = 0$ and $T = 1$, ie $$ t\to t'(t) = t + \epsilon $$ Then, we have $$ \delta S = \int_{t_1}^{t_2} \frac{\partial L}{\partial t} \,dt $$ If $L$ has no explicit time dependence, the result follows with $K = 0$.
Second, we can choose $X = \dot q$ and $T = 0$, ie $$ x\to x'(x,t) = x + \epsilon \dot q(t) $$ Then, we have \begin{align} \delta S &= \int_{t_1}^{t_2}\left[ \frac{\partial L}{\partial x}\dot q + \frac{\partial L}{\partial v} \ddot q \right]\,dt \\&= \int_{t_1}^{t_2}\left[ \frac{d}{dt} L(q,\dot q,t) - \frac{\partial L}{\partial t} \right]\,dt \end{align}
If $L$ has no explicit time dependence, we again arrive at our conservation law, but this time with $$ K(t) = L(q(t), \dot q(t), t) $$
Your approach follows this second path. However, as $K \not= 0$, we're dealing only with a quasi-symmetry of the action, so your assumption $\delta S = 0$ was not warranted.