Equivalent Lagrangians
Two Lagrangians which yield the same equations of motion are called equivalent Lagrangians. It follows that Lagrangians which differ by a constant are equivalent Lagrangians (because the equations of motion derived from them turn out to be the same). In your case, $\mathcal L-\mathcal L_1=mv_0 ^2/2=\rm constant$, thus $\mathcal L$ and $\mathcal L_1$ are equivalent. For a general case, if two Lagrangians ($\mathcal L$ and $\mathcal L'$) are equivalent then
$$\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial(\mathcal L-\mathcal L')}{\partial \dot q_i}\right)=0 \quad \text{and} \quad \frac{\mathrm d (\mathcal L-\mathcal L')}{\mathrm dq_i}=0 \qquad \forall \:\: i\tag{1}$$
Derivation
Let's prove that the Lagrangians in any two inertial reference frames will always be equivalent. Let the Lagrangian in the ground frame be
$$\mathcal L=T-V=\sum_i^n \frac 1 2 m_i |\mathbf v_i|^2-\sum_j^m\left(\sum_i^n V_i(q_j)\right)$$
where $m_i$ and $\mathbf v_i$ are the mass and the velocity of the $i$-th particle, and $V_i$ is the potential of the $i$-th particle dependent of the coordinate $q_j$. Now let's move to a refrence frame moving with a velocity $\mathbf v$, then the new Lagrangian is
$$\mathcal L'=T'-V'=\sum_i^n \frac 1 2 m_i |\mathbf v_i-\mathbf v|^2-\sum_j^m\left(\sum_i^n V_i(q)\right)$$
Therefore,
$$\mathcal L-\mathcal L'=\sum_i^n \frac 1 2 m_i |\mathbf v_i|^2-\sum_i^n \frac 1 2 m_i |\mathbf v_i-\mathbf v|^2$$
Now using the fact that $|\mathbf v_i-\mathbf v|^2=|\mathbf v_i|^2+|\mathbf v|^2-2\mathbf v_i\cdot \mathbf v$, we get
\begin{align}
\mathcal L-\mathcal L'&=\sum_i^n \frac 1 2 m_i (|\mathbf v_i|^2-(|\mathbf v_i|^2+|\mathbf v|^2-2\mathbf v_i\cdot \mathbf v))\\
&= \sum_i^n m_i (\mathbf v_i\cdot \mathbf v)-\sum_i^n \frac 1 2 m_i |\mathbf v|^2\\
&= \left(\sum_i^n m_i \mathbf v_i\right)\cdot \mathbf v-\sum_i^n \frac 1 2 m_i |\mathbf v|^2
\end{align}
Now we'll use the fact that $\displaystyle\sum_i^n m_i \mathbf v_i=M\mathbf v_{\rm COM}$, where $\displaystyle M=\sum_i^n m_i$ and $\mathbf v_{\rm COM}$ is the velocity of the center of mass. Thus,
\begin{align}
\mathcal L-\mathcal L'&=M(\mathbf v_{\rm COM}\cdot \mathbf v)-\sum_i^n \frac 1 2 m_i |\mathbf v|^2
\end{align}
Let the coordinate along the direction of $\mathbf v$ be $q_k$, then $\mathbf v_{\rm COM}\cdot\mathbf v=\dot{q_k}|\mathbf v|$. Substituting this, we get
\begin{align}
\mathcal L-\mathcal L'&=M\dot{q_k}|\mathbf v|-\sum_i^n \frac 1 2 m_i |\mathbf v|^2\\
\mathcal L-\mathcal L'&=M\dot{q_k}|\mathbf v|-\rm constant
\end{align}
Now testing the conditions in equation $(1)$, we get:
\begin{align}
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial(\mathcal L-\mathcal L')}{\partial \dot q_k}\right)&=\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial(M\dot{q_k}|\mathbf v|-\rm constant)}{\partial \dot q_k}\right)\\
&=\frac{\mathrm d}{\mathrm dt}(M |\mathbf v|)\\
&=0\\
&\text{and}\\
\frac{\mathrm d(\mathcal L-\mathcal L')}{\mathrm d q_k}&=\frac{\mathrm d(M\dot{q_k}|\mathbf v|-\rm constant)}{\mathrm d q_k}\\
&=0
\end{align}
As you see that the conditions in equation $(1)$ hold true, which implies that both the Lagrangians are equivalent. This could also be trivially proven by showing the invariance of equations of motions under changing from one inertial reference frame to another.
Conclusion
Your instructor's statement (as you stated) is completely incorrect. However, it is correct to say that Lagrangians obtained in different inertial frames of reference are always equivalent.
