The problem you pose stems from an implicit false dichotomy introduced from the various meanings of "solve".
Any equation can be solved, whether polynomial, linear or transcendental. The algebraic process of solving usually involves algebraic transformations which preserve or do not preserve the equivalence of the solutions to the original one. When it does not preserve it, care is taken to examine superfluous or extraneous solution.
The last step, not always but usually, involves inverting a given function. The problem is that the domain of functions which are algebraically invertible (i.e. the inverse has a closed form which allows direct calculations) is somewhat more limited than the domain of functions which are generally invertible.
Suppose, for example, I have reached the step and need to solve:
$$e^z+z^2+1/z+z^8=0$$
Solving for $z$ amounts to inverting $f(z)=e^z+z^2+1/z+z^8$. The existence of such an inverse is always guaranteed by the Lagrange Inversion Theorem, so away from critical points, there always exists an $f^{-1}(w)$, such that:
$$w=f(z)\Leftrightarrow$$
$$z=f^{-1}(w)$$
Call this $f^{-1}$ "myCrazyInverse(w)". It exists and is well defined in a neighborhood of a good solution, provided the solution is not a critical point of $f$.
This function can now be approximated, using numerical methods, provided it is fairly smooth locally and therefore the original equation can be solved to the user's $\epsilon$ satisfaction.
The inverse of this particular example either has a closed form, or it doesn't. If it does, we dump this example to the list of "alebraically" solvable examples and are done with, since the inversion can be done manually. If it doesn't, it may take a while for mathematicians to hold-on to this function in the mathematical memory of things, depending on usage.
For example, it's fairly useless to classify the inverse of this $f$ in the official archives of mathematical functions, because its use has been very rare. Have you seen it before? Chances are nil. It is a very rare kind of function (the inverse) and functions are usually classified in order of some use.
The most notable (and fairly late) example you may probably already be familiar with, is the Lambert $W$ function. Its use has become somewhat "standard" lately, not because it just "solves" the transcendental equation $z\exp(z)=w$, rather because mathematicians have encountered it often-enough in various branches of science, such as astronomy, quantum physics and optics, away from mathematics themselves, which guaranteed a fairly frequent usage.
If you didn't know its properties, it would be fairly useless, too. The following implication (as an algebraic step) is actually valid:
$$z\cdot \exp(z)=y\Leftrightarrow z=W(y)$$
but away from knowing actual ways to calculate values $W(y)$, it is as moot a step as the inversion above with "mycrazyinverse".
It just so happens that this function is of use in many fields, hence the step of the inversion of $z\exp(z)=y$ has been included in the allowed algebraic transformations which solve all the related transcendental equations and provide a "closed form" solution.
Even under the above notes, many mathematicians still doubt the inclusion of this inversion step, given the multivalued nature of (most crazy) inverses and I actually sympathize with them. In view of the multivalued nature of some maps, it can be argued that such steps are valid algebraically only under specific qualifications, which preclude semantic mathematical nonsense, such as:
$$z\cdot\exp(z)=y\Leftrightarrow z=W(y)$$
The correct step is qualified by the branch index only and denotes an equivalence of sets:
$$z\cdot\exp(z)=y\Leftrightarrow z=W(k,y),\,\,\,k\in\mathbb{Z}$$
You can wittness the resulting confusion in answers which use the Lambert $W$ function to solve equations, as a wrong answer: $x=W(whatever)$. Whenever this function solves an equation, it always specifies infinitely many solutions, whether real or complex, because of the branch index. It's never a single value, whether the teacher wants only real solutions or not.
Addednum (for clarification for #2)
For some reason I didn't see that last question, so I am adding a reply here.
The Lambert $W$ function enjoys certain algebraic properties, which allow it to solve many transcendental equations, such as close variants of the original $z\cdot \exp(z)=y$, like $z^3\cdot \exp(z)=y$.
All these are exactly the equations which through some valid algebraic manipulations can be eventually be brought to the form:
$$f(z)\cdot\exp(f(z))=y\,\,\,\,(1)$$
for some $f$ and some $y$, since the solutions can then be described as:
$$f_k(z)=W(k,y)\Leftrightarrow z_k=f^{-1}(W(k,y)),\,\,\,\,k\in\mathbb{Z}$$
Maple contains an algorithm which does exactly that in each given case, but the algorithm's success alone is not a proof, because the algorithm has to stop sometime.
Fortunately, correct algebraic syntax makes for a solvable algebra/group (by definition), so you can be sure that if some equation can be factored into form (1), it will be factored in $O(g(n))$ time, where the latter is the factoring time of the algorithm and this gives you deterministic certainty with a very small fail rate $\epsilon(g(n))>0$.
Now push $\epsilon\to 0$ and you've proved it.
