Convolution by IIR filter, a case where circular convolution is allowed? The question is, how to attain the frequency domain representation of any signal (which is finite, non-unit length) convolved with an IIR filter. 

The correct answer in my opinion is that there is in fact no answer, which might seem very strange at first. But if we work through the math and use the property of circular convolution we of course will still get an answer.

So let's say our signal is simple discrete step function of the length N-1, starting at t=0, the frequency domain representation is then: 

$$F(\omega)=\large e^{j \omega\frac {N-1}{2}}$$

To make the point clearer let's say the IIR filter is of the form. $h(n) = \cos(\omega_0 n)$, which converter to frequency domain is equal to:

$$
H(\omega)=
\begin{cases} 
1, &amp; \omega = \omega_0
\\ 0, &amp; \text{otherwise}
\end{cases}
$$

Calculating the frequency representation of the convolution can be done using the formula $$Y(\omega)=F(\omega) H(\omega)$$

Where $Y(\omega)$ is the frequency domain representation of the convolution. However, there is a problem, convolving these functions necessarily results in time domain aliasing, as there is no possible way of having zeroes near the edges. And so, the aliasing occurs, thus the result is:

$$
F(\omega)  G(\omega)=
\begin{cases} 
\large e^{j \omega\frac {N-1}{2}}, &amp; \omega = \omega_0
\\ 0, &amp; \text{otherwise}
\end{cases}
$$

In time domain the result is either sinusoidal wave or nothingness, depending on the frequency of the sine wave and width of the pulse we started with.

So continuing this line of thought, sine waves in LTI systems have linear response because of circular convolution (as without the fold back the sine wave would distort). This means that LTI systems properties are a completely theoretical construct, which we of course already knew since infinite sine waves with no end or beginning don't exist in real world. Either way, my line of thought is that you can't convolve an infinite signal using the equation, as no amount of infinities makes a signal fold back onto itself, as using the math will do (admittedly, I may be completely wrong on this). I will be trying to solve the equation for a case where the sine wave is casual next...

A frequency representation of a particular interval can of course be attained (which conviniently cuts the impulse response to a FIR, removing the funny things caused by infinities). The problem also disappears if the filter settles to zero after infinite time, which probably satisfies the definition of IIR, but is of no interest for the purpose of the discussion.

Am I correct on this or does letting the aliasing occur lead to the correct answer for the convolution functions in cases of infinite time functions? If that is the case, what is the rationale behind it?
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.