Consider the equation
$$g(t) = \int_a^b K(t,s)f(s) ds $$
where $g$ and the kernel $K$ are known and $f$ is to be determined. Suppose that the equation has a solution. Under what conditions on the kernel $K$ (could be special cases or general results) is the solution $f(\cdot)$ to the above equation unique?
For example, as described on Wikipedia, there is a unique solution when $K(t,s) = K(t-s)$ and the limits $a$ and $b$ are at $-\infty$ and $+\infty$ respectively.
Here are some thoughts. I am not sure if there are positive results out there, but basically, the answer seems to be that solutions are almost never unique, even when $K$ has the special form $K(s,t) = K(s-t)$.
Let's denote the operator in question by $T$. So $T: L^2(a,b) \to L^2(a,b)$ is defined by $$Tf(t) = \int_a^b \! K(t,s) f(s), ds.$$
When $K(t,s)$ is of the special form $K(t-s)$, then $T$ will be unitarily equivalent to a multiplication operator on $L^2(\mathbb{R})$ or $\ell^2$ by considering the problem in Fourier space. In this case theorems of Paley-Wiener type show that if $K$ is very regular, the null space $N(T)$ is large so that solutions are highly non-unique. To be more explicit, let us consider an example at the extreme end of this idea. Suppose for a second that $(a,b) = \mathbb{R}$ and that $K(t) \in L^1(\mathbb{R})$ is integrable so that the corresponding convolution operator makes sense. Then if $F$ is the Fourier transform we have for $f \in L^2(\mathbb{R})$ $$FTF^{-1}(f) = F(K \ast F^{-1}f)= \hat{K}f$$ so that $T$ is equivalent to multiplication by $\hat{K}$. If $K(t)$ were a function which is the restriction to $\mathbb{R}$ of an entire function $\tilde{K}: \mathbb{C} \to \mathbb{C}$ satisfying an bound of the form $$\tilde{K}(z) \le Ae^{M|z|}$$ (so that $K$ is very regular) then $\hat{K}$ has compact support (see for example Complex Analysis by Stein & Shakarchi). This means that $N(T)$ is infinite-dimensional and thus for every solution to an equation of the form $g = Tf$ there exist infinitely many other linearly independent solutions $g = T(f + h)$.
Results in the negative direction can also be obtained under much less stringent conditions when, for example, $(a,b) = (-\pi, \pi)$. In this case the "convolution" operator $T$ is still equivalent to a multiplication operator on $\ell^2$ by the sequence $(\hat{K}(k))_{k \in \mathbb{Z}}$. The high-brow way of seeing this is by noticing that functions on $(-\pi,\pi)$ are really just functions on the circle group, but you can also just extend functions in question periodically and use the definitions.
So now $T$ is injective if and only if $\hat{K}(k) \neq 0$ for all $k \in \mathbb{Z}$. This is almost never the case! In fact, for each $k$, the subspace of $L^2(-\pi, \pi)$ with $\hat{K}(k) = 0$ is infinitely dimensional of co-dimension $1$. For every trigonometric polynomial, $\hat{K}(k) = 0$ for all but finitely many $k$. Even innocent functions such as $K(t) = \frac{\pi}{2} - |t|$ satisfy $\hat{K}(k) = 0$ for infinitely many $k$.
In fact, even when $\hat{K}(k) \neq 0$ for every $k \in \mathbb{Z}$, The Riemann-Lebesgue Lemma tell us that $\hat{K}(k) \to 0$ as $k \to \infty$. So even when the inverse exists, it is never bounded.
To get general positive results, the equation probably has to be interpreted in some distributional sense. For example statements such as "if $K$ is a $\delta$-potential, then $T$ is the identity operator" can be formalized in an appropriate framework. But I'm going to take the liberty of assuming that this is probably not what you had in mind when you posted the question.
I think you would have much better luck in searching for uniqueness results of equations of the second kind. In this case you are examining the null space of $I + T$, which tends to be finite-dimensional, or even trivial, by Fredholm Theory.