Convolution by IIR filter, a case where circular convolution is allowed?
To answer your first question formulated in the first sentence: if $X(\omega)$ is the Fourier transform of the input signal, and $H(\omega)$ is the frequency response of the IIR filter, the Fourier transform ("frequency domain representation" as you call it) of the output signal is simply $Y(\omega)=X(\omega)H(\omega)$. This is valid in continuous time as well as in discrete time if you use the discrete-time Fourier transform (DTFT) (and not the DFT/FFT).

In the (discrete) time domain, you generally have an infinite convolution sum:

$$y[n]=\sum_{k=-\infty}^{\infty}x[k]h[n-k]\tag{1}$$

where $x[n]$ is the input sequence, and $h[n]$ is the impulse response of the filter. If either $x[n]$ or $h[n]$ (or both) are of finite length, the sum in $(1)$ becomes a finite sum.

If you manage to clarify the rest of your question, I might be able to add more relevant information to this answer.

If I may guess what confuses you, it may be that you think in terms of the DFT, which can only handle sequences of finite length (and may cause circular convolution artefacts). However, you can implement an IIR filter directly (and exactly, if we ignore quantization effects for the moment) in the time domain. In the general case, the DTFT of the output signal can only be computed approximately. However, for finite length input signals fed into an IIR filter, we can obtain an exact expression for the DTFT of the input signal as well as for the frequency response of the IIR filter. The Fourier transform of the output signal is then simply the multiplication of the two.