Coming at this from a slightly more abstract point of view than the other answers, this is an application of the multiplicative cancellation law (over the rationals if $a$, $b$, and $c$ are all integers; or over the reals if $a$, $b$ and $c$ are real numbers; or over the complex numbers; or whatever...). Specifically, in this context, the cancellation law says:
Let $q$, $r$, and $s$ be rational numbers (or real numbers, or complex numbers) with $q\ne 0$. Then $q\cdot r = q\cdot s$ if and only if $r=s$.
Note that it is quite important here that $q \ne 0$. If $q = 0$, then both $r$ and $s$ may be chosen freely, and no cancellation is possible.
Taking $q = \frac{1}{b}$ (assuming that $b\ne 0$, $q$ is well-defined and we automatically have $q\ne 0$), $r = x$, and $s = c$, the multiplicative cancellation law gives us
$$
\frac{x}{b} = \frac{c}{b}
\iff \frac{1}{b} \cdot x = \frac{1}{b} \cdot c
\iff x = c. $$
Because we are using the multiplicative cancellation law, the process is called cancelling or cancelling out the common factor (in this case, we are cancelling a factor of $\frac{1}{b}$). Indeed, I think that a properly rigorous reading of this step would be "We cancel out a common factor of $\frac{1}{b}$."
It might also be reasonable to say that "We cancel the common factors from the denominators," or more simply "We cancel the denominators." That said, because I am kind of pedantic, I would be a little hesitant to say that anything is being done to $b$. We aren't really cancelling a factor of $b$, but rather a factor of $\frac{1}{b}$.