Inspired by this proof in MathWorld, I rewrote the proof in terms of the Ramanujan theta function.
Define the function $$M(c)=\prod_{n = 1}^{\infty}(1 +a^{n}b^{n-1}c)\left(1+\frac{a^{n-1}b^{n}}{c}\right)\tag1$$
then $$M(abc)=\prod_{n = 1}^{\infty}(1 +a^{n+1}b^{n}c)\left(1+\frac{a^{n-2}b^{n-1}}{c}\right)\tag2$$
$$M(abc)=(1+a^2bc)\left(1+\frac{1}{ac}\right)(1+a^{3}b^{2}c)\left(1+\frac{b}{c}\right)(1+a^{4}b^{3}c)\left(1+\frac{ab^{2}}{c}\right)\cdots$$
$$M(c)=(1+ac)\left(1+\frac{b}{c}\right)(1+a^{2}bc)\left(1+\frac{ab^{2}}{c}\right)(1+a^{3}b^{2}c)\left(1+\frac{a^{2}b^{3}}{c}\right)\cdots$$
Taking $$\frac{M(abc)}{M(c)}=\left(1+\frac{1}{ac}\right)\left(\frac{1}{1+ac}\right)=\frac{1}{ac}$$
yields the following relation $$M(c)=acM(abc).$$
Now define $$N(c)=M(c)\prod_{n = 1}^{\infty}(1 -(ab)^{n}).$$
Then $$N(abc)=M(abc)\prod_{n = 1}^{\infty}(1 -(ab)^{n})$$
which becomes $$N(c)=acN(abc).$$
Now expand $N(c)$ in a Laurent series $$N(c)=\sum_{n=-\infty}^{\infty}u_{n}c^{n}.$$
Using the fundamental relation, we have $$\sum_{n=-\infty}^{\infty}u_{n}c^{n}=ac\sum_{n=-\infty}^{\infty}u_{n}(abc)^{n}$$
$$=\sum_{n=-\infty}^{\infty}u_{n}a^{n+1}b^{n}c^{n+1}$$
$$=\sum_{n=-\infty}^{\infty}u_{n-1}a^{n}b^{n-1}c^{n}.$$
Which leads to the recurrence relation $$u_{n}=u_{n-1}a^{n}b^{n-1}.$$
$$u_{1}=u_{0}a,$$ $$u_{2}=u_{1}a^{2}b=u_{0}a^{3}b,$$ $$u_{3}=u_{2}a^{3}b^{2}=u_{0}a^{6}b^{3},$$ $$u_{4}=u_{3}a^{4}b^{3}=u_{0}a^{10}b^{6}.$$
Which in general form is $$u_{n}=u_{0}a^{n(n+1)/2}b^{n(n-1)/2}.$$
Now substituting back into the original Laurent series, we obtain $$N(c)=u_{0}\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}c^{n}.$$
It can easily be shown that $u_{0}=1$, so that we have the Jacobi triple product in terms of the Ramanujan theta function $$N(c)=\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}c^{n}.$$
Q:Is this generalisation of the proof correct?
Revised: 25 April 2020
Following Somos' critique, I searched around a couple of papers and finally came across the following on wikipedia
Let us set $a=b=q=e^{2\pi i \tau}$ and $c=1$ and show that the following numerator and denominator
$\frac{1}{u_{0}(e^{2\pi i n^2 \tau})}=\frac{\sum_{n=-\infty}^{\infty}e^{2\pi i n^2 \tau}}{\prod_{k = 1}^{\infty}(1-e^{4\pi i k \tau})\left(1+e^{2\pi i (2k-1) \tau}\right)^2}$
are weight 1/2 modular functions under the transformation $\tau = \frac{-1}{4 \tau}$
Since a modular function is 1-periodic $f(q+1)=f(q)$ and bounded on the upper half plane $\tau \to i \infty$, it ought to be constant according to Louiville's theorem, therefore the quotient has to be constant such that $u_{0}(q)=u_{0}(0)=1$