In principle, yes. In practice, for small quantities of salt, no. This is a very simple experiment to perform.

If you're interested in the theory, this is mostly chemistry. The elevation in boiling temperature caused by adding the salt [can be expressed as][1]:
$$\Delta T=Kbi$$
where $K\approx 0.5 \,\mathrm{K \,kg \,mol^{-1}}$ for water is called the ebullioscopic constant, $i\approx 2$ for salt (in water $\mathrm{NaCl}$ dissociates into two) and $b$ is the molality, that is the number of moles of salt per mass of water.

Now the amount of heat necessary to reach a certain temperature is proportional to the difference between the final and initial temperature. Assuming a constant heat source, the time it takes to heat to a certain temperature is proportional to the amount of heat needed. Therefore if $T_0$ is the temperature at which water boils, $T_i$ is the initial temperature, $t_w$ is the time it takes for water only to boil, and $t_w + \Delta t$ is the time it takes for water + salt to boil we get:
$$\frac{t_w + \Delta t}{t_w}=\frac{T_0+\Delta T-T_i}{T_0-T_i} \implies \frac{\Delta t}{t_w}=\frac{\Delta T}{T_0-T_i} $$
So now suppose we start at $T_i=20^\circ \mathrm{C}$ and add say $10 \mathrm{g}$ of salt; we can use [this online calculator][2] to obtain the number of moles, which in this case is $0.17\mathrm{mol}$. Suppose we use $1\mathrm{kg}$ of water. Of course $T_0=100^\circ \mathrm{C}$ so we get:

$$\frac{\Delta t}{t_w}=\frac{Kbi}{T_0-T_i} \approx 0.2\%$$
so not very relevant. Suppose instead we put in $0.5\mathrm{kg}$ of salt. Then the corresponding value is:
$$\frac{\Delta t}{t_w}\approx 10\%$$
which becomes relevant, but is still not a lot.
  [1]: https://en.wikipedia.org/wiki/Boiling-point_elevation
  [2]: https://www.convertunits.com/from/moles+NaCl/to/grams