frequency - Under what conditions does jw equal the laplace variable s in an electrical circuit?

Imagine I've got an electromagnetic actuator and solved the underlying differential equations. I got an solution for sinusodially excitation, as follows (just an example, no physical backround):

$$G(\mathrm{j}\omega) = \frac{I(\mathrm{j}\omega)}{U(\mathrm{j}\omega)} = \frac{1}{1 + \mathrm{j}\omega T}$$

Now I just set

$$\mathrm{j}\omega = s, \quad s \rightarrow \text{Laplace Variable}$$

to get

$$G(s) = \frac{1}{1 + s T}$$

and I claim that this is the transfer function for the electrical circuit feeding the actuator. I'm sure the results I get are correct, but I was told I can't just set $$\mathrm{j}\omega = s$$ without further explication.

So I wonder what are the conditions to do what I've done, how can I proof that both equations are correct and explain that in a correct mathematical way? I feel that everybody (in literature) either just does it or avoids the issue.

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Further explanation:

Above system G(jw) shows a low-pass behavior in the frequency domain, means for sinusodially voltage excitation $$ i(\mathrm{j}\omega) = G(\mathrm{j}\omega) \cdot u(\mathrm{j}\omega) = \frac{1}{1+\mathrm{j}\omega T} \cdot u(\mathrm{j}\omega)$$ I get a higher damping the higher the frequency of the excitation.

But we also know that a transfer function $$ i(s) = G(s) \cdot u(s) = \frac{1}{1+s T} \cdot u(s)$$ excited by a unitary step $$u(s) = \frac{u_0}{s}$$ will also lead to a valid solution, namely the typical exponential $$ i(t) = u_0 \cdot (1- e^{-t/T})$$ PT1-behavior.

So what is the condition, that a system which is valid in the frequency domain for harmonic signals in steady-state, is also valid in the Laplace-domain by setting $$ \mathrm{j}\omega = s$$ for the non-steady-state e.g. for excitation with a unitary step?