Is it correct to assume my resolution is $f_s/N$?
Yes, but it's not useful.
You won't clearly see the frequency of the sine in your FFT, because your signal model is, as far as your description goes,
$$
s(t) = A_0\sin(2\pi f_\text{sine}t + \varphi_0) \cdot e^{-\tau t}.\label1\tag1
$$
Pay attention to the multiplication of the sine with the exponential envelope! In frequency domain, that's a convolution of the two Fourier transforms of the sine and the exponential. With the latter not being band-limited, you can't even sample this 100% correctly, but more importantly, the things that are "easy" in time domain become smeared and unclear in frequency domain. The FFT is making your life harder here! It's the wrong tool.
Instead, consider Eq. $\eqref{1}$ simply as an equation with four unknowns, and solve it by using four or more samples, in time domain. It's not hard to do it generally (it becomes slightly more complicated when you add noise, but let's ignore that for the time being).
First, I'd take two samples that are simple a sampling period $T_s$ apart, and divide them:
\begin{align}
\frac{s(t_0)}{s(t_0+T_s)} &=
\frac
{A_0\sin(2\pi f_\text{sine}t_0 + \varphi_0) \cdot e^{-\tau t_0}}
{A_0\sin(2\pi f_\text{sine}(t_0+T_s) + \varphi_0) \cdot e^{-\tau (t_0+T_s)}}\\
&=\frac
{\sin(2\pi f_\text{sine}t_0 + \varphi_0) }
{\sin(2\pi f_\text{sine}(t_0+T_s) + \varphi_0)}
\cdot e^{-\tau [t_0-(t_0+T_s)]}\\
%
&=\frac
{\sin(2\pi f_\text{sine}t_0 + \varphi_0) }
{\sin(2\pi f_\text{sine}(t_0+T_s) + \varphi_0)}
\cdot e^{\tau T_s}\\
\end{align}
Do the same for the sample one period earlier (at $t=t_0-T_s$), and then divide the results. You'll see that the result doesn't contain your exponential term anymore, the $\sin(2\pi f_\text{sine}t_0 + \varphi_0) $ itself cancels, and you're just left with an equation that contains the product of two sines. From there it's applying a formula to simplify that product of sines and your observed sample values $s(t_0\pm T_s)$; and you'll see how you can solve this!
