This conjecture is equivalent to the following
$$\frac{q}{(1-q)^2}\sum_{n=0}^\infty(-q)^n \frac{(q;q^2){}_n(-q^2;q^2){}_n}{(q^3;q^2){}_n^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)},\tag{1}$$
and after using the identity pointed out by M. Somos (with $q$ relaced by $-q$)
$$
\frac{(q;-q)_\infty}{(-q;-q)_\infty}\sum_{n=-\infty}^\infty\frac{q^{n^2}}{(1-q^{2 n-1})^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)}.\tag{1a}
$$
Initial proof of $(1a)$ utilized the identity
$$
\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)-1}=\frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s}),\tag{2}
$$
which turned out to be another conjecture due to Michael Somos. So that initial proof was not self sufficient. Self sufficient proofs of $(1a)$ and $(2)$ are given below.
Proof of $(1a)$.
By shifting the summation of variables so that all summations start from $0$ and by elementary rearrangements of terms we get
\begin{align}
\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)-1}&=2\sum_{r,s,t\ge 0}q^{\frac{r^2+3r}{2}+s^2+2s+2rs+t(r+1+2s)}-\sum_{1\le r\le t}q^{t^2-(r^2-r)-1}\\
&=\left(\sum_{r,s,t\ge 0}+\sum_{r,s,t< 0}\right)q^{\frac{r^2+3r}{2}+s^2+2s+2rs+t(r+1+2s)}\\
&=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-q^{r+1+2s}}.
\end{align}
It turns out that it is easier to work with a more general sum
$$
g(z,q)=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+2s}}.
$$
Following the general method outlined in the article by E. Mortenson we derive the functional equation satisfied by $g(z,q)$:
\begin{align}
0&=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right){q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}\\
&=\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+1+2s}}({1-z\,q^{r+1+2s}})\\
&=g(qz,q)-qz\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+5r}{2}+s^2+4s+2rs}}{1-z\,q^{r+1+2s}}\\
&=g(qz,q)-\frac{z}{q}\left(\sum_{r\ge 1,s\ge 0}-\sum_{r\le 0,s<0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z\,q^{r+2s}}\\
&=g(qz,q)-\frac{z}{q}g(z,q)+\frac{z}{q^2}j\left(-q,q^2\right) m\left(-{z }/{q},q^2,-q\right),\tag{3}
\end{align}
where in the last line we used notations
$$
J_m=(q^m;q^m)_\infty,\quad j(x,q)=(q;q)_\infty(x;q)_\infty(q/x;q)_\infty,\quad m(x,q,z)=\frac1{j(z,q)}\sum_{n=-\infty}^\infty\frac{(-z)^nq^{n(n-1)/2}}{1-xzq^{n-1}}.\tag{4}
$$
Since the Appell-Lerch sum $m(x,q,z)$ satisfies the equation (eq. 2.2c in Mortenson's paper)
$$
m(qx,q,z)=1-xm(x,q,z),\tag{5}
$$
one can show by elementary calculation that the function
$$
g_1(z,q)=-\frac{z}{q^3}j\left(-q,q^2\right) m\left(-{z }/{q^2},q^2,-q\right) m\left(-{z}/{q},q^2,-q\right)
$$
satisfies the same functional equation $(3)$ as $g(z,q)$. Now consider the function
$$
h(z,q)=g(z,q)-g_1(z,q),
$$
for which we have the functional equation
$$
h(qz,q)=\frac{z}{q}h(z,q).\tag{6}
$$
From the definition of $g(z,q)$ it is easy to see that it has simple poles at $z_0=q^n,~n\in\mathbb{Z}$ with residues being equal to finite sums of powers of $q$, i.e. polynomials in $q$. For example for $n\ge 1$ the residue is
$$
\sum_{r+2s=n}q^{\frac{r^2-r}{2}+s^2+2rs}.
$$
When $m\left(-{z }/{q^2},q^2,-q\right) $ is singular then $m\left(-{z }/{q},q^2,-q\right) $ is not and vice versa. So from the definition of $m(x,q,z)$ one can see that $g_1(z,q)$ has the same set of simple poles as $g(z,q)$. To find residues one can use the identity $m(-1,q^2,-q)=0$ and the functional equation $(5)$. The result is that $h(z,q)$ is analytic. Analytic function with functional equation $(6)$ is $0$ (for details one can consult Mortenson's paper, section 5). So we proved that
$$
\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-z q^{r+2s}}=-\frac{z}{q^3}j\left(-q,q^2\right) m\left(-{z }/{q^2},q^2,-q\right) m\left(-{z}/{q},q^2,-q\right).\tag{7}
$$
We want to calculate this expression when $z=q$. It is known that $m\left(-{z }/{q^2},q^2,-q\right)$ is singular at $z=q$ and $m\left(-{z }/{q},q^2,-q\right)=0$ at $z=q$. So one needs to resolve this ambiguity by applying L'Hôpital's rule:
\begin{align}
&\lim_{z\to q}m\left(-{z }/{q^2},q^2,-q\right)m\left(-{z }/{q},q^2,-q\right)\\
&=\frac{q}{j\left(-q,q^2\right) }\lim_{z\to q}\frac{m\left(-{z }/{q},q^2,-q\right)}{1-z/q}\\
&=\frac{q}{j\left(-q,q^2\right)}\frac{\frac d{dz}m\left(-{z }/{q},q^2,-q\right){\Large|_{\small{z=q}}}}{-1/q}\\
&=-\frac{q^2}{j^2\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}q^{2n-2}}{(1-q^{2n-1})^2}\\
&=-\frac{q}{j^2\left(-q,q^2\right)}\left(\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}-\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{1-q^{2n-1}}\right)\\
&=-\frac{q}{j^2\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}.
\end{align}
So when $z\to q$ eq. $(7)$ becomes
$$
\left(\sum_{r,s\ge 0}-\sum_{r,s< 0}\right)\frac{q^{\frac{r^2+3r}{2}+s^2+2s+2rs}}{1-q^{r+1+2s}}=\frac{1}{qj\left(-q,q^2\right)}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}\\
=\frac{(-q;-q)_\infty}{q(q;-q)_\infty}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2}.\Box
$$
Proof of $(2)$.
Here it is proved that
$$
\frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s})=\frac{(-q;-q)_\infty}{q(q;-q)_\infty}\sum_{n=-\infty}^{\infty}\frac{q^{n^2}}{(1-q^{2n-1})^2},(*)
$$
which together with $(1a)$ will imply $(2)$.
The sums of the type $\sum_{r,s,t\ge 0}q^{rs+st+tr+r+t+s}$ have been studied by Mortenson, Corollary 1.2, where he proved that
\begin{align}
\sum_{r,s,t\ge 0}q^{rs+st+tr}x^{r+t+s}+&\sum_{r,s,t< 0}q^{rs+st+tr}x^{r+t+s}=\\
&=3\frac{J_1^3j(x^2,q)}{j(x,q)^2}m(-q/x,q^2,qx^2)-2\frac{J_1^3J_2^3j(x^2,q^2)^3}{j(x,q)^3j(-x,q^2)^3}\tag{8}
\end{align}
with the notations $(4)$.
Since the difference of two Appel-Lerch functions $m(x,q,z)$ with different $z$ is a quotient of theta functions (eq. 2.6 in Mortenson's paper) we can write in $(8)$
$$
\small{m(-q/x,q^2,qx^2)=m\left(-{q}/{x},q^2,z\right)+z\frac{J_2^3 j\left({q x^2}/{z},q^2\right) j\left(-xzq^2,q^2\right)}{j\left(q x^2,q^2\right) j\left(-xq^2,q^2\right) j\left(z,q^2\right) j\left(-{q z}/{x},q^2\right)}.}\tag{8a}
$$
We want to calculate the triple sum in $(8),(8a)$ when $x=q$. Note that
$$
\sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\sum_{r,s,t< 0}q^{rs+st+tr}q^{r+t+s}.
$$
However $j(q,q)=0$ and since $j(x,q)$ appears in the denominator at the RHS of $(3)$ one needs to use L'Hôpital's rule to calculate the limit $x\to q$ with the help of formulas $\frac d{dx}j(x/q,q){\Large|_{\small{x=q}}}=-\frac1q J_1^3$ and $j(q^nx,q)=(-1/x)^nq^{-n(n-1)/2}j(x,q)$.
First, we put $z=-q$ and take the limit $x\to q$ in $(8),(8a)$ to obtain
\begin{align}
&\sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}\\
&=\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}-\frac{8 J_2^{12}}{j\left(-q,q^2\right)^3 J_1^6}-3\frac{j\left(-q^2,q^2\right) J_2^6}{q^2j\left(-q,q^2\right) j\left(q^3,q^2\right) j\left(-q^3,q^2\right) j\left(q,q^2\right)}.
\end{align}
One can show that
$$
\frac{8 J_2^{12}}{j\left(-q,q^2\right)^3 J_1^6}+3\frac{j\left(-q^2,q^2\right) J_2^6}{q^2j\left(-q,q^2\right) j\left(q^3,q^2\right) j\left(-q^3,q^2\right) j\left(q,q^2\right)}=2\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3}.
$$
So
$$
\sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}-2\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3}.\tag{9}
$$
Similarly, the choice $z=q$ in $(8),(8a)$ leads to
$$
\sum_{r,s,t\ge 0}q^{rs+st+tr}q^{r+t+s}=\frac{3 }{q j\left(q,q^2\right)}\sum _{r=-\infty}^\infty \frac{(-1)^rq^{r^2}}{\left(1+q^{2 r-1}\right)^2}+4\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3},\tag{10}
$$
and after changing sign $q\to-q$:
$$
\sum_{r,s,t\ge 0}(-q)^{rs+st+tr+r+t+s}=-\frac{3 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}+4\frac{\left(q^4;q^4\right){}_{\infty }^3}{\left(q^2;q^4\right){}_{\infty }^3},\tag{10a}
$$
Combine $(9)$ and $(10a)$ to get
$$
\frac13\sum_{r,s,t\ge 0}(2q^{rs+st+tr+r+t+s}+(-q)^{rs+st+tr+r+t+s})=\frac{1 }{q j\left(-q,q^2\right)}\sum _{r=-\infty}^\infty \frac{q^{r^2}}{\left(1-q^{2 r-1}\right)^2}.
$$
Now observe that $\frac{(q;-q)_{\infty } j\left(-q,q^2\right)}{(-q;-q)_{\infty }}=1$ to complete the proof of $(*)$.$~ \Box$
