Let $B_t$ denote the $t$-value of a riskless money market account in which 1 unit of currency has been invested at time $t=0$. We have $B_t = e^{rt}$, where $B_t$ solves
$$dB_t = B_trdt,\quad B_0=1 \tag{0} $$
If $S_t$ represents the $t$-value of a tradable asset, then in the absence of arbitrage opportunity we must have that (fundamental theorem of asset pricing):
$$ \frac{S_t}{B_t} \text{ is a } \Bbb{Q} \text{-martingale} $$
in other words
\begin{align}
d\left(\frac{S_t}{B_t}\right) &= \frac{dS_t}{B_t} - \frac{S_t}{B_t^2} dB_t + 0 \tag{Itô} \\ &= \frac{1}{B_t} \left( dS_t - S_t r dt \right) \tag{using $(0)$} \\
&= \dots dW_t^\Bbb{Q} \tag{1}
\end{align}
by the martingale representation theorem.
Note that for $(1)$ to hold it is necessary to have:
$$ dS_t = \color{blue}{rS_t} dt + \dots dW_t^\Bbb{Q} $$
This is the equivalent martingale measure requirement you're looking for.
Now, to reach that $rS_t$ drift under $\Bbb{Q}$, you shall not use the same Girsanov kernel depending on whether $S_t$ follows a GBM or an ABM under $\Bbb{P}$ but that's another question.
In the first case you'll have: $$dS_t = \alpha S_t dt + \sigma S_t dW_t^\Bbb{P} \to dS_t = \color{blue}{rS_t} dt + \sigma S_t dW_t^\Bbb{Q} $$
with $$\left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert_{\mathcal{F}_t} = \mathcal{E}\left[ -\lambda W_t^\Bbb{P}\right], \quad \lambda=\frac{\alpha-r}{\sigma}$$
In the second:
$$dS_t = \alpha dt + \sigma dW_t^\Bbb{P} \to dS_t = \color{blue}{rS_t} dt + \sigma dW_t^\Bbb{Q} $$
with $$\left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert_{\mathcal{F}_t} = \mathcal{E}\left[ -\lambda W_t^\Bbb{P}\right], \quad \lambda=\frac{\alpha-rS_t}{\sigma}$$
See this paper if you want more mathematical details.
