Note 1. First of all, this is a very nice question, congratulations! You're very right to question those "weirdnesses" that arise from complex numbers, and not just accept them as magical. Keep it up!
[...] solutions that aren't visible by graphing this equation on your calculator
I would like to start this answer with a surprise: it is possible to see it graphically!! Your calculator plotted a 2D graph, right? Your calculator was thinking in real numbers only. The "complete" graph, taking into account complex numbers, would have to be 4D! If it was possible to plot a 4D graph, it would be possible to see the two solutions! Since 4D graphs are clearly impossible, though, there is a "hack" using two 3D graphs:
I won't be very detailed, because this is only a motivation for the actual answer, but basically you must visualize, in each 3D graph, a horizontal plane (on zero) intersecting the plotted surfaces. Then, you must look for a point that lies in both intersections. And only the points $(0,1)$ and $(0,-1)$ will satisfy this condition (these points are precisely $i$ and $-i$, respectively)!
[...] create your own solutions to this problem
Some teachers introduce the concept of complex numbers by saying something on the lines of "until now we believed that there is no solution to $x^2 = -1$, but let's define a number, and call it $i$, such that $i^2 = -1$", and see what happens".
This makes it look like "the problem has no solutions, but we will create a solution nevertheless". You are totally right to seek a similar thing: can we create a solution to $x \times 0 = 1$, and call it $k$? This is why I don't like this approach, it is very misleading.
But there is a problem. A more "correct" approach would require a lot more abstract, mathematical thinking, which is inappropriate for several audiences. Not inappropriate for you, though! I will try to give you a start. If you feel motivated, you are very welcome to post further questions in this site and continue learning.
A better approach to complex numbers (optional part)
You are used to manipulate real numbers. We have the numbers, and some operations we can do to them, especially adding and multiplying. Also, there operations have some nice properties. For example, both are commutative, right? It would be weird if addition and multiplication weren't commutative. In algebra, we work with other objects, that are not the real numbers, but "behave similarly", in the sense that we can also do things similar to adding and multiplying them. When we have a set of objects like this, we call this set a field. So the real numbers, alongside the usual addition and multiplication operations, are an example of field. The rational numbers are also a field.
Now, if we take any set of objects whatsoever, and manage to define two operations with them that behave similarly to the addition and multiplication of real numbers, we have a field! I think this is the best way of thinking about complex numbers. What we do is:
Definition of a "complex number": a "complex number" is a pair of real numbers. The first is called the "real part" and the second is called the "imaginary part" (but for now these are just random names, that could be "tomato" and "potato" instead).
Definition of the addition-like operation: the sum of $(a,b)$ with $(c,d)$ is $(a+b,c+d)$.
Definition of the multiplication-like operation: the product of $(a,b)$ with $(c,d)$ is $(ac-bd, ad+bc)$.
And with these definitions, it turns out that we also have a field!! That's why most of the behavior of complex numbers are very similar to the real numbers: both are fields.
Now, something amazing happened: what happens if I multiply the complex $(0,1)$ by itself? The result, according to the rules already defined above, is $(-1,0)$.
Now we can define a new notation. Instead of writing $(a,b)$ to represent this pair of real numbers, we will write $a+bi$ instead. Then, we suddenly realize that, from what was said above, it is a fact, not a definition, that $i^2 = -1$.
Your actual question
Now addressing your specific question about $x \times 0 = 1$: No, it is not possible. As I said above, the sets of rational numbers, real numbers and complex numbers are all fields. And it is a mathematical fact (that I won't prove here for brevity) that in any field whatsoever, anything times zero is equal to zero. If you want to define a solution to $x \times 0 = 1$, your extended set of numbers can't be a field. And this is very bad: you technically can do it, but you won't be able to manipulate the numbers as you're used to. It will break almost all arithmetic rules you learned since you was a kid.