Assuming the existence of solutions in solving exercises

A sizeable chunk of my first calculus course at university comprised of learning techniques to evaluate limits, such as this simple example, evaluating the limit: $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7}.$$

A typical solution would be to identify that for $x \neq 7$, $$\frac{x^2 -8x + 7}{x-7} = x-1,$$ so $$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} = \lim_{x \to 7} x-1 = 6.$$

In my eyes, we have shown that if the limit exists, its value must be $6$. We have not shown that the limit exists in the first place and is equal to $6$, since we have presupposed the existence of the limit when writing

$$\lim_{x \to 7} \frac{x^2 -8x + 7}{x-7} = \lim_{x \to 7} x-1,$$

since the existence of both objects on either side of an equality is a necessary condition for the equality to be true (right?).

My main questions are: do such methods of evaluation serve as evidence that these limits in fact exist in the first place, or do they only tell us what the limit ought to be, and the only way we can be sure is to formally prove it using the $\epsilon$-$\delta$ definition? Is this case similar to "finding" the derivatives of functions?

There's no logical problem with this argument. The expressions $$ \frac{x^2 -8x + 7}{x-7} \text{ and } x-1 $$ are equal when $x \ne 7$, so the first expression has a limit at $7$ if and only if the second does. There is no need to assume the existence of the limit in advance.

Whether or not you need the $\epsilon - \delta$ argument to find the the limit of $x-1$ depends on the level of rigor your instructor requires.

(There are other situations where a correct argument does have the form

the limit is such and such provided the limit exists

usually followed by a separate proof that there is a limit.)

Note that when we find the limit of a function at a point the value of the function at that point is not important.

In your example the function $f(x) =\frac {x^2-8x+7}{x-7}$ and $g(x)=x-1$ have the same values at every point except at $x=7$ thus they have the same limits at that point.

Since $g(x)$ has a limit of $6$ at $x=7$ so does $f(x)$.