Your function for the pressure (which you denote with s(r, t)) is correct but let us express it in the complex domain (we will take the real part at the end of our calculations). It becomes (Please note that I have also rearranged the terms in the exponent)
$$s(r, t) = \frac{A}{r} e^{j(\omega t - k r)} ~ (1)$$
Now, from Euler's equation (momentum equation) we know that
$$\nabla s = - \rho \frac{\partial u}{\partial t} ~ (2)$$
where $\rho$ is the medium density (we assume that the perturbed density is identical to the static one) and $u$ is the particle velocity.
In spherical coordinates, after omitting the radial components (due to symmetry, there is no change in any of the radial components and the derivative with respect to them is identically zero) Euler's equation becomes
$$\frac{\partial s}{\partial r} = - \rho \frac{\partial u}{\partial t} ~ (3)$$
By combining equations (1) and (3) we get
$$-\rho \frac{\partial}{\partial t} \left[u e^{j \left( \omega t - k r \right)} \right] = \frac{\partial}{\partial r} \left[ \frac{A}{r} e^{j \left( \omega t - k r \right)} \right] \implies
-j \rho \omega u = -\frac{\left(A j k r + A \right)}{r^{2}} e^{j \left(\omega t - k r \right)} \implies \\ \implies -j \rho \omega u = -\frac{A \left(1 + j k r \right)}{r^{2}} e^{j \left(\omega t - k r \right)} \implies -j \rho \omega u = -\frac{\left(1 + j k r \right)}{r} \frac{A}{r} e^{j \left(\omega t - k r \right)} \implies \\ -j \rho \omega u = -\frac{\left(1 + j k r \right)}{r} s \implies -j \rho \omega u = -\frac{s + s j k r}{r} \implies u = \frac{s + s j k r}{j \rho \omega r} \implies \\ \implies u = \frac{s}{j \rho \omega r} + \frac{s k}{\rho \omega} \implies u = \frac{s k}{\rho \omega} \left( 1 + \frac{1}{j k r} \right) ~ (4)$$
Now, by noting that $k = \frac{\omega}{c} \implies c = \frac{\omega}{k}$ you write the last part of (4) as
$$u = \frac{s}{\rho c} \left( 1 + \frac{1}{j k r} \right) ~ (5)$$
Such a lengthy derivation it is but we are going to need it. Before moving forward to look into how this can help us to find out how to calculate the intensity it's worth noting that the relation of pressure and particle velocity is frequency ("hidden" in the wavenumber $k$) and distance dependent. The further away you go from the source, or the higher the frequency becomes, the more in-phase those two quantities are.
Next, based on what I assume to be considered known, we state that the intensity is the product of sound pressure and particle velocity. That is
$$I = p u ~ (6)$$
Since we use complex notation but we are only interested in the real part of the intensity we get (please refer to some textbook for a more in-depth explanation of why this is true)
$$I_{r} = \frac{1}{2} \Re \left[ p u^*\right] ~ (7)$$
where the subscript $r$ denotes that this is the real part of the quantity, $\Re$ denotes the real part and $[ ~ \cdot ~ ]^{*}$ denotes complex conjugation. Thus, by using equations (1) and (5) into (7) we get
$$I_{r} = \frac{1}{2} \Re \left[ \frac{A e^{j \left(\omega t - k r \right)}}{r} \frac{A^{*} e^{-j \left(\omega t - k r \right)}}{\rho c r} \right] \implies I_{r} = \frac{1}{2} \Re \left[ \frac{A A^{*} e^{j \left(\omega t - k r \right)} e^{-j \left(\omega t - k r \right)}}{\rho c r^{2}} \right] \implies \\ \implies I_{r} = \frac{\left| A \right|^{2}}{2 \rho c r^{2}} \implies I_{r} = \frac{\left| s \right|^{2}}{2 \rho c} ~ (8)$$
where in the last part of (8) $I_{r}$ is valid only for time averages of harmonic functions over some multiple of the period (usually root-mean-square values are used instead for real life applications).
For more information on sound intensity you can refer to some well known textbooks such as Fundamentals of Acoustics by Kinsler et al. or (maybe even better) Sound Intensity by Frank Fahy. Please note that those are mere suggestions based on my personal preferences and experience and they are not meant to act as some promotion of said resources. Additionally, please bear in mind that you could very well find equivalent information on other "general acoustics" textbooks.
Best Answer
It is a typo and the negative sign disappears because it is a cosine function, $s(x,t)=s_{\rm max}\cos(kx∓ωt+ϕ)$, which is being differentiated.
$y$ is the displacement of the centre of mass of a parcel of the medium from its equilibrium position at a fixed position $x$ from the origin.
So this displacement relates to the movement of the medium not the movement of the wave profile (a "photograph" of the wave at one instant of time).