I think calling some operations bi-directional is not the clearest way to put it. What I think you mean is that while for all functions $f$ and $x,y$ in the domain of $f,$ $$x=y \implies f(x)=f(y),$$ but $$f(x)=f(y) \;\not\!\!\!\implies x=y.$$ I think a better name would be invertible operations. If your students have been introduced to the concept of function inverses (I would expect so in a calculus class), then operation inverses should be an easy extension.
The inverse of adding one is subtracting one. The inverse of multiplying by three is dividing by three. However, not all operations have inverses--at least, not always.
Operations with restricted domains eliminate solutions
$$x^2 = x$$
$$\frac{x^{2}}{x} = \frac{x}{x}$$
$$x = 1$$
Most students will be able to look at the original equation and see that $x=0$ is a solution as well. Ask them what happened to zero in the worked out solution. One way to explain this is that since division by zero is not defined, dividing by $x$ eliminates zero as a possible value for $x$. Multiplying by zero does not have an inverse operation because dividing by zero is undefined.
Operations that are not one-to-one fabricate solutions
$$x = 3$$
$$x^2 = 9$$
$$x^2 - 9 = 0$$
$$(x - 3)(x + 3) = 0$$
$$x = 3 \quad \textrm{or} \quad x = -3$$
Where did the extra solution come from? You could say that since $x^2 = (-x)^2,$ squaring both sides of an equation mixes up answers with different signs. Squaring has no inverse for negative inputs.
Operations that aren't quite inverses
$$x^2 = 9$$
$$\sqrt{x^2} = \sqrt{9}$$
$$x = 3$$
Where did the solution $x = -3$ go? The square root operation only gives positive results, so a square root is not an inverse operation of squaring when the input is negative. This is related to the previous example. A non one-to-one function cannot have an inverse for all inputs. One can use $\sqrt{x^2} = |x|$ to get all of the correct answers.
Other thoughts
I wonder if the above is related to the trouble I and others had with the proving trigonometric identities back in high school. When problems were of the form:
$$\textrm{Prove that: } \textrm{A mess of trig functions} = \textrm{Another mess of trig functions}$$
a common strategy was to manipulate the equation until it resulted in something like $1 = 1$ and declare that, "Since the original equation is equivalent to $1=1,$ it must be true." This is obviously fallacious since one could prove that $\sin\theta=\cos\theta$ just by multiplying both sides of the equation by zero to get the obviously true $0=0.$ However, if all of the operations were invertible, then we would have a valid proof just be reversing the order of the lines: start with $1=1$ and end up with the identity.
