There are two good answers I can see, I will explain both.
If we define the functions $\ln x$ and $e^x$ for real inputs and outputs, there is a problem with $\ln x$ when $x$ is negative. This can be seen as follows: graphing the exponential function $y=e^x$, one can see that it does not pass beneath the $x$-axis. This means that $e^x> 0$ for all $x\in \mathbb{R}$. It follows that there is no real number $x$ such that $e^x=-7$ and thus while we would like for $e^{\ln(-7)}=-7$ for the reason you gave, there is no real number that would play the right role for $\ln(-7)$. More generally, we fail to find a good candidate (in the real numbers, see below) for what $\ln x$ should be if $x$ is negative. For this reason, one sometimes declares that $\ln x$ is defined only for $x>0$. This is the reason you see people here saying that the statement is false. You could use this reasoning to conclude that rather than true or false the statement is ill-defined (which I think would be a little more accurate).
Now there is another context in which there is a number $x$ such that $e^x=-7$, namely if we allow complex numbers. If you are not familiar with them, you can revisit this part after you learn them. If you are familiar, the answer goes like this. The exponential function $e^z$ can be defined for complex numbers $z=x+iy$:
$$e^z=e^x(\cos y+i\sin y)$$
to understand this, look up Euler's formula (but ignore stuff about his other formula $V-E+F=2$). This definition allows $e^z$ to equal any non-zero complex number, therefore we can solve $e^z=-7$ if $z$ can be complex. This leads to the definition that Metehan Turan gives in their answer. In this case, $$\ln(-7)=\ln 7+i\pi$$
Plugging this in, we see that
\begin{align*}
e^{\ln 7+i\pi}&=e^{\ln 7}(\cos \pi+i\sin \pi)\\
&=-7
\end{align*}
so, you can say that the statement is true on this reading.
*It should be noted that there is more than one complex number that is a reasonable value for the natural logarithm of a given non-zero complex number. We say that it is multivalued or that it is not uniquely defined.
Addendum
You may ask why people would bother with the real-number-only version of the logarithm and the answer lies largely with the fact that the complex logarithm is not uniquely defined, and, perhaps worse, whichever definition is given it is not continuous! This is pretty bad, but it is a foundational element of the beautiful subject of complex analysis. A deep study of that subject will clarify all of these questions.
Best Answer
The question's been basically answered already, but here's another way to look at it without dividing at all, if you're afraid of division (like I sometimes am). It is more formal in a way, but will not let you down if in doubt.
$$f(x)g(x)=f(x)g'(x)$$ is the same as $$f(x)\Big(g(x)-g'(x)\Big)=0$$ which is true whenever (that is, it's the same as)
$$f(x)=0\;\text{or}\;g(x)=g'(x)$$
because a product is zero when either multiple is zero.
The next step in this formal approach is to see carefully where the universal quantifier is:
$$\forall x:\big(f(x)=0\;\text{or}\;g(x)=g'(x)\big)$$
For every $x$, you know that either of $f(x)=0$ or $g(x)=g'(x)$ is true, but you do not know which of the two. And this is the (formal) reason you cannot say anything about $g$ at points where $f(x)=0$.
To conclude something about $g$ at some $x_0$, you need to additionally assume that the left operand of "or" is false - that is, $f(x_0)\neq0$. And note how we're reasoning here only pointwise - for each $x$ a different choice of the two conditions might be true. That's why @Jules talks about the different intervals in his answer.