Are all complex functions onto?

Since I feel like your question is trying to address finding some kind of generalisation of the fundamental theorem of algebra (which can be rephrased as saying any nonconstant complex polynomial is surjective), I think one of the best things you can get is Picard's little theorem.

First, let me mention that continuous complex functions are not necessarily surjective, even if they aren't constant: for example, the absolute value function $|\cdot|:\Bbb C\to\Bbb R_{\geq0}\subset\Bbb C$ is certainly continuous, but is also certainly not surjective. Therefore, just having continuity is not enough, so if we want to remedy this by making the functions in question look "more like polynomials", then we ought to make them smoother; that is, (complex) differentiable.

It turns out being complex differentiable is quite a bit to ask: unlike in the real case, a function that is complex differentiable will automatically be analytic; that is, it will have a Taylor series expansion at the point where it is differentiable. Therefore, differentiable complex functions can be thought of as "infinite degree polynomials", and we can go back to the question of: does the fundamental theorem of algebra somehow generalise to this setting?

Cutting to the chase, one place we might end up is Picard's little theorem, which says that if our complex function is differentiable everywhere and also not a constant, then its image will be just about surjective; that is, its image will be $\Bbb C$ except possibly a single point. Therefore, given a complex function $f:\Bbb C\to\Bbb C$ that is differentiable, you will be able to solve $f(z)=a$ for all $a\in\Bbb C$ with at most one exception.

For the record, an example of an entire function whose image is missing a point would be the exponential map $\exp:\Bbb C\to\Bbb C$, whose image is $\Bbb C\setminus\{0\}$.

As I understand it, your question is raising an analogy with the fact that $x^2 + 1 = 0$ has no solutions over $\Bbb{R}$, and we need to "invent" a new number $i = \sqrt{-1}$ (that is, adjoin an element with the appropriate algebraic properties to the field $\Bbb{R}$). This gives us the field $\Bbb{C}$. So your question is, do we have to keep "inventing" new numbers [adjoining new elements] $j, k, l, ...$ indefinitely to $\Bbb{C}$ to solve ever broader classes of equations?

As far as polynomials go, the answer is "no". $\Bbb{C}$ is algebraically closed: if you give me any polynomial $p(x) \in \Bbb{C}[x]$ with complex coefficients, I can give you a complex number $z$ for which $p(z) = 0$. $\Bbb{C}$ is also closed under field operations: for any two complex numbers $a_1 + b_1i$, $a_2 + b_2i$, their sum, difference, product, and quotient can all be written in the form $a + bi$ as well (so we don't need to keep inventing new symbols $j, k, l, ...$ to handle basic field operations either).

It's even true that complex numbers have logarithms and roots, although these are generally multivalued, and cannot be defined on all of $\Bbb{C}$ because of singularity issues at the origin. I can also raise a complex number to a complex power, and my result still is in $\Bbb{C}$. Basically, once you add $i$ to the field $\Bbb{R}$, your work of adding elements is complete.

While it is not completely clear what you mean by "operations within the complex numbers", it almost sounds that you are alluding to the fundamental Theorem of Algebra. This Theorem says that any polynomial in $\mathbb{C}$ factors completely over $\mathbb{C}$. So, all the roots "stay inside the complex numbers". If you are familiar with Field Theory, this is equivalent to their being no algebraic extensions of $\mathbb{C}$.

However, you second thought still does not hold completely in the complex numbers. You cannot just invert something like $z^2$ because it is not a bijective function on $\mathbb{C}$. To define $\sqrt{z}$ you can write this as $z^{1/2} = e^{1/2\log z}$ where $\log z$ is a branch of the logarithm. Remember that the logarithm is not analytic on all of $\mathbb{C}$ and you you will need to take a branch cut.

If such an inverse exists, then it is necessarily continuous and and analytic where the original function is nonzero (albeit, these are true in the non-complex case as well).