Let us assume that $a,b,c$ are real numbers.
The integral in question equals
\begin{equation}
I=I_1+I_2,
\end{equation}
where
\begin{equation}
I_1:=\int_0^2 dt\; t^{-1/2}e^{-1/t} J_a(t)J_b(t),\quad I_2:=\int_2^\infty dt\; t^{-1/2}e^{-1/t} J_a(t)J_b(t),
\end{equation}
\begin{equation}
J_a(t):=\int_0^\infty\frac{dx}{(1+x)^a(1+x+t)^c}.
\end{equation}
Clearly, if $I<\infty$, then $J_a(t)<\infty$ for almost all $t>0$, whence $a+c>1$. Similarly, it is necessary that $b+c>1$. So, without loss of generality
\begin{equation}
a+c>1,\quad b+c>1. \tag{1}
\end{equation}

So, for $t\in(0,2]$,
\begin{equation}
J_a(t)\asymp\int_0^\infty\frac{dx}{(1+x)^a(1+x)^c}\asymp1,
\end{equation}
whence one always has
\begin{equation}
I_1\asymp1<\infty,
\end{equation}
given the necessary condition (1).

For $t>2$,
\begin{align}
J_a(t)&\asymp\int_0^1\frac{dx}{t^c}+\int_1^t\frac{dx}{x^a t^c}+\int_t^\infty\frac{dx}{x^a x^c} \\
&\asymp t^{-c}+t^{-c}\int_1^t\frac{dx}{x^a}+t^{-c+1-a},
\end{align}
whence
\begin{equation}
J_a(t)\asymp
\begin{cases}
t^{(1-a)_+-c}&\text{ if } a\ne1,\\
t^{-c}\ln t&\text{ if } a=1.
\end{cases}
\end{equation}
Since $e^{-1/t}\asymp1$ for $t>2$, it follows that
\begin{equation}
I_2<\infty\iff (1-a)_+-c+(1-b)_+-c-1/2<-1
\iff c>\tfrac14+(1-a)_+ +(1-b)_+.
\end{equation}
In particular, the latter condition implies $c>(1-a)_+\ge1-a$ and similarly $c>1-b$, so that the necessary condition (1) holds.

Thus, for $I<\infty$ it is necessary and sufficient that
\begin{equation}
c>\tfrac14+(1-a)_+ +(1-b)_+.
\end{equation}