Sometimes just using the integral form of error function can save you.
$\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}
\text{erf}\biggl(\dfrac{b}{\sqrt{x}}\biggr)~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^\frac{b}{\sqrt{x}}e^{-u^2}~du~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^1e^{-\left(\frac{bu}{\sqrt{x}}\right)^2}~d\biggl(\dfrac{bu}{\sqrt{x}}\biggr)~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^t\dfrac{1}{x}e^{-\frac{a^2}{x}}\int_0^1\dfrac{b}{\sqrt{x}}e^{-\frac{b^2u^2}{x}}~du~dx$
$=\dfrac{2b}{\sqrt{\pi}}\int_0^t\int_0^1\dfrac{e^{-\frac{b^2u^2+a^2}{x}}}{x\sqrt{x}}du~dx$
$=\dfrac{2b}{\sqrt{\pi}}\int_0^1\int_0^t\dfrac{e^{-\frac{b^2u^2+a^2}{x}}}{x\sqrt{x}}dx~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^t-e^{-\frac{b^2u^2+a^2}{x}}~d\biggl(\dfrac{1}{\sqrt{x}}\biggr)~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_\infty^\frac{1}{\sqrt{t}}-e^{-(b^2u^2+a^2)x^2}~dx~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_\frac{1}{\sqrt{t}}^\infty e^{-(b^2u^2+a^2)x^2}~dx~du$
$=\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^\infty e^{-(b^2u^2+a^2)x^2}~dx~du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\int_0^\frac{1}{\sqrt{t}}e^{-(b^2u^2+a^2)x^2}~dx~du$
$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\biggl[\sum\limits_{n=0}^\infty\dfrac{(-1)^n(b^2u^2+a^2)^nx^{2n+1}}{n!(2n+1)}\biggr]_0^\frac{1}{\sqrt{t}}~du$
$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\sum\limits_{n=0}^\infty\dfrac{(-1)^n(b^2u^2+a^2)^n}{t^{n+\frac{1}{2}}n!(2n+1)}du$
$=2b\int_0^1\dfrac{1}{\sqrt{b^2u^2+a^2}}du-\dfrac{4b}{\sqrt{\pi}}\int_0^1\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^na^{2n-2k}b^{2k}u^{2k}}{t^{n+\frac{1}{2}}n!(2n+1)}du$
$=2b\biggl[\dfrac{\ln\left(b^2u+b\sqrt{b^2u^2+a^2}\right)}{b}\biggr]_0^1-\dfrac{4b}{\sqrt{\pi}}\biggl[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^{2n-2k}b^{2k}u^{2k+1}}{t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}\biggr]_0^1$
$=2\ln\left(b^2+b\sqrt{a^2+b^2}\right)-2\ln(|a|b)-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n4a^{2n-2k}b^{2k+1}}{\sqrt{\pi}t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}$
$\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\text{erfc}\left(\sqrt{\dfrac{a^2+x^2}{t}}\right)~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_\sqrt{\frac{a^2+x^2}{t}}^\infty e^{-u^2}~du~dx$
$=\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_0^\infty e^{-u^2}~du~dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\int_0^\sqrt{\frac{a^2+x^2}{t}}e^{-u^2}~du~dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\biggl[\sum\limits_{n=0}^\infty\dfrac{(-1)^nu^{2n+1}}{n!(2n+1)}\biggr]_0^\sqrt{\frac{a^2+x^2}{t}}~dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a^2+x^2)^{n+\frac{1}{2}}}{t^{n+\frac{1}{2}}n!(2n+1)}dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a^2+x^2)^n}{t^{n+\frac{1}{2}}n!(2n+1)}dx$
$=\int_0^b\dfrac{1}{\sqrt{a^2+x^2}}dx-\dfrac{2}{\sqrt{\pi}}\int_0^b\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^na^{2n-2k}x^{2k}}{t^{n+\frac{1}{2}}n!(2n+1)}dx$
$=\left[\ln\left(x+\sqrt{a^2+x^2}\right)\right]_0^b-\dfrac{2}{\sqrt{\pi}}\biggl[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^{2n-2k}x^{2k+1}}{t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}\biggr]_0^b$
$=\ln\left(b+\sqrt{a^2+b^2}\right)-\ln|a|-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2a^{2n-2k}b^{2k+1}}{\sqrt{\pi}t^{n+\frac{1}{2}}k!(n-k)!(2n+1)(2k+1)}$
Your integral admits a closed form.
Theorem. Let $a$ and $b$ be any real numbers. Then
$$\int_0^{\infty }\!\! e^{\large-\left(\frac{a-\log x}{b}\right)^2}\!\! \left(\text{erf}\left(\frac{a-\log x}{b}\right)+1\right) \mathrm{d}x=\sqrt{\pi}\:|b|\:e^{{\large {a+\frac{b^2}{4}}}}\left(1-\text{erf}\!\left(\! \frac{\sqrt{2}\:b}{4}\!\right)\! \right) \tag1$$
where $\displaystyle \text{erf}(z)=\frac{2}{\sqrt{\pi }}\int _0^ze^{\large-t^2}\mathrm{d}t.$
Proof. Let's denote the left hand side of $(1)$ by $I(a,b)$.
By the change of variable $$\displaystyle U:=\frac{a-\log x}{b}, \quad dx=-b\:e^a \:e^{-b U}dU,$$ we have
$$
\begin{align}
I(a,b)&=|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \left(\text{erf}\left(U\right)+1\right) \mathrm{d}U\\\\
&=|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \text{erf}(U)\: \mathrm{d}U+|b|\:e^a \int_{-\infty }^{+\infty } e^{\large-U^2-bU} \mathrm{d}U \tag2
\end{align}
$$ Clearly, by the gaussian integral:
$$
\int_{-\infty }^{+\infty } e^{\large-U^2-bU} \mathrm{d}U =e^{\large b^2/4}\int_{-\infty }^{+\infty } e^{\large-(U+b/2)^2} \mathrm{d}U =\sqrt{\pi}\:e^{ b^2/4} .\tag3
$$ Set
$$
f(b):=\int_{-\infty }^{+\infty } e^{\large-U^2-bU} \text{erf}(U)\: \mathrm{d}U \tag4
$$ and observe that $$ f(0)=\frac{\sqrt{\pi}}{4}\left[(\text{erf}(U))^2\right]_{-\infty }^{+\infty }=0. \tag5$$ Differentiating $(4)$ with respect to $b$ and performing an integration by parts gives
$$
\begin{align}
f'(b)&=-\int_{-\infty }^{+\infty } U\:e^{\large-U^2-bU} \text{erf}(U)\:\mathrm{d}U\\\\
f'(b)&=\left[\frac 12 e^{\large-U^2}\left(e^{\large-bU} \text{erf}(U)\right)\right]_{-\infty }^{+\infty }-\frac 12\int_{-\infty }^{+\infty } e^{\large-U^2} \left(-b\:e^{\large-bU} \text{erf}(U)+\frac{2}{\sqrt{\pi }}e^{\large-bU}e^{\large-U^2}\right)\mathrm{d}U\\\\
f'(b)&=\frac{b}{2}f(b)-\frac{1}{\sqrt{\pi }}\int_{-\infty }^{+\infty } e^{\large-2U^2-bU}\mathrm{d}U
\end{align}
$$ or
$$
f'(b)=\frac{b}{2}f(b)-\frac{\sqrt{2}}{2}\:e^{\large b^2/8}. \tag6
$$ We classically solve the ODE $(6)$, using $(5)$, to obtain
$$
f(b)=-\sqrt{\pi }\:e^{\large b^2/4}\text{erf}\!\left(\! \frac{\sqrt{2}\:b}{4}\!\right) \tag7
$$ then plugging $(7)$, $(4)$ and $(3)$ into $(2)$ gives $(1)$ as desired.
Best Answer
The result reads: \begin{eqnarray} &&\int\limits_0^\infty x e^{-a x^2-b x} \text{erf}(c x+d) dx= \frac{e^{\frac{b^2}{4 a}}}{4 a} \left( \right.\\ && \frac{2 b \left(\arctan\left(\frac{\sqrt{a} (b+2 c d)}{2 a d-b c}\right)+\arctan\left(\frac{2 \sqrt{a} d}{b}\right)-\arctan\left(\frac{c}{\sqrt{a}}\right)\right)}{\sqrt{\pi } \sqrt{a}} +\\ && 2 e^{-\frac{b^2}{4 a}} \text{erf}(d)-\frac{\sqrt{\pi } b \text{erf}\left(\frac{2 a d-b c}{2 \sqrt{a \left(a+c^2\right)}}\right)}{\sqrt{a}}+\frac{2 c e^{-\frac{(b c-2 a d)^2}{4 a \left(a+c^2\right)}} \text{erfc}\left(\frac{b+2 c d}{2 \sqrt{a+c^2}}\right)}{\sqrt{a+c^2}} +\\ && -\frac{4 \sqrt{\pi } b \left(T\left(\frac{2 a d-b c}{\sqrt{2} \sqrt{a \left(a+c^2\right)}},\frac{\sqrt{a} (b+2 c d)}{2 a d-b c}\right)+T\left(\frac{b}{\sqrt{2} \sqrt{a}},\frac{2 \sqrt{a} d}{b}\right)\right)}{\sqrt{a}}\\ &&\left.\right) \end{eqnarray} where $T(\cdot,\cdot)$ is the Owen's t function https://en.wikipedia.org/wiki/Owen%27s_T_function .