I believe the property you're referring to is

$$\mathcal{L}\left\{ f(t-\tau) \right\} = e^{-s\tau}\mathcal{L}\left\{f(t)\right\}$$

This means that your transfer function does not really follow that rule, and does not really have a time delay:

$$H(s) = \frac{s}{as^2 + bs + c + de^{-sT}}$$

Solving the inverse Laplace transform for this function will most likely involve numerical methods. What is possible, is to have something like this:

$$H(s) = \frac{s + de^{-sT}}{as^2 + bs + c}$$

Because it can be separated in:

$$H(s) = \frac{s}{as^2 + bs + c} + e^{-sT}\frac{d}{as^2 + bs + c}$$

You can see that the second term also can be separated in partial fractions, like the first term. The poles of the denominator will then say something about the stability of the transfer function.

Appendix

It can be noted that \$e^{-sT}\$ can be expanded in its Taylor series:

$$e^{z} = \sum_{n=0}^{+\infty} \frac{z^n}{n!}$$

So this means your example transfer function can be written as

$$\begin{align} H(s) &= \frac{s}{as^2 + bs + c + d\cdot (\sum_{n=0}^{+\infty} \frac{(-sT)^n}{n!})} \\ &= \frac{s}{\sum_{n=0}^{+\infty} A_ns^n} \end{align}$$

Where \$A_0 = c + d\$, \$A_1 = b - d\cdot T\$, \$A_2 = a + d\frac{T^2}{2}\$, \$A_n = d\frac{(-T)^n}{n!}, \forall n>2\$.

So this kind of transfer function has "poles" all over the place. This is to illustrate that you definitely can't deal with this like a regular second-order transfer function.

I believe the property you're referring to is

$$\mathcal{L}\left\{ f(t-\tau) \right\} = e^{-s\tau}\mathcal{L}\left\{f(t)\right\}$$

This means that your transfer function does not really follow that rule, and does not really have a time delay:

$$H(s) = \frac{s}{as^2 + bs + c + de^{-sT}}$$

Solving the inverse Laplace transform for this function will most likely involve numerical methods. What is possible, is to have something like this:

$$H(s) = \frac{s + de^{-sT}}{as^2 + bs + c}$$

Because it can be separated in:

$$H(s) = \frac{s}{as^2 + bs + c} + e^{-sT}\frac{d}{as^2 + bs + c}$$

You can see that the second term also can be separated in partial fractions, like the first term. The poles of the denominator will then say something about the stability of the transfer function.

Stability in general

The formula for the inverse Laplace transform is

$$f(t) = \frac{1}{2\pi j}\lim_{T\to \infty}\int_{\sigma-jT}^{\sigma+jT} F(s)e^{st}ds$$

where \$\sigma\$ is chosen to be greater than all singularities of \$F(s)\$ on the complex plane (in our case such that it includes all poles).

This integral is equivalently solved by the Cauchy residue theorem:

$$f(t) = \mathcal{L}^{-1}\left\{ F(s) \right\} = \sum_{all\ poles\ of\ F(s)} Res\left[ F(s)e^{st} \right]$$ Remember, this is general, ie. it always works out for a transfer function with whatever singularities!

So it doesn't matter whether or not the poles come from a polynomial or from a transcendental function. As long as the singularities are all in the left-half plane, their residue will always contain an exponential that decays to 0 rather than infinity. Any singularities in the RHP will always lead to an exponential that explodes.

Appendix

It can be noted that \$e^{-sT}\$ can be expanded in its Taylor series:

$$e^{z} = \sum_{n=0}^{+\infty} \frac{z^n}{n!}$$

So this means your example transfer function can be written as

$$\begin{align} H(s) &= \frac{s}{as^2 + bs + c + d\cdot (\sum_{n=0}^{+\infty} \frac{(-sT)^n}{n!})} \\ &= \frac{s}{\sum_{n=0}^{+\infty} A_ns^n} \end{align}$$

Where \$A_0 = c + d\$, \$A_1 = b - d\cdot T\$, \$A_2 = a + d\frac{T^2}{2}\$, \$A_n = d\frac{(-T)^n}{n!}, \forall n>2\$.

So this kind of transfer function has "poles" all over the place. This is to illustrate that you definitely can't deal with this like a regular second-order transfer function.

Appendix

The residue of a (simple) pole is

$$Res_{s=a}\left[ H(s) \right] = \lim_{s\to a} (s-a)F(s)$$

For poles with multiplicity \$n\$:

$$Res_{s=a}\left[ H(s) \right] = \frac{1}{(n-1)!} \lim_{s\to a} \frac{d^{n-1}}{ds^{n-1}}\left( (s-a)^n H(s) \right)$$