How about this:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\sum_{n=0}^{\infty}\frac{2}{\sqrt{\pi}}\int_n^{\infty}e^{-t^2}\, dt
$$
In this sum of integrals, the interval $[0,1)$ will be counted only once, in the $n = 0$ term. The interval $[1,2)$ will be counted twice, in the $n = 0$ and $n = 1$ terms. And so on. This means we can write:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\frac{2}{\sqrt{\pi}}\int_0^{\infty}\, \lfloor t+1\rfloor \,e^{-t^2}\, dt
$$
I don't know if this is the sort of alternative integral representation you were looking for.
Edited to add two other ways to write this expression:
First way: The quantity $\lfloor t+1\rfloor$ can be written as
$$
\lfloor t+1\rfloor \;=\; (t+1) \;-\; S(t)\, .
$$
where $S(t)$ is a sawtooth wave of period $1$ with minimum value $0$ and a maximum value $1$. One way we could write this sawtooth is as $S(t) \,=\, t\;\mathrm{mod}\;1$. Substituting this expression for $\lfloor t+1\rfloor$ into the integral above and using the fact that $\int_0^{\infty} dt \, (t+1)\,e^{-t^2} = (1+\sqrt{\pi})/2$ yields
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\;
\frac{2}{\sqrt{\pi}}\left[\frac{1}{2}(1+\sqrt{\pi}) \;-\; \int_0^{\infty}dt\, S(t)\, e^{-t^2}\right].
$$
Evaluating this numerically in Mathematica with 20 digits of precision yields $1.16200283409802758182$, which is greater than Mathematica's estimate for the original sum by roughly $3.8\times {10}^{-6}$. Close enough for the vagaries of numerically evaluating weird sums and integrals.
Second way: Poisson's summation formula states that if $f(x)$ is a function and
$$
\hat{f}(q) \;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, f(x)
$$
is its Fourier transform, then
$$
\sum_{n = -\infty}^{+\infty} f(n) \;=\; \sum_{n=-\infty}^{+\infty} \hat{f}(2\pi n)\, .
$$
Define the even function $f(x) = \mathrm{erfc}(|x|)$. Since $f$ is even, we can write the original sum as
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n) \;=\; \frac{1}{2}f(0) \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} f(n)
\;=\; \frac{1}{2} \,+\,\frac{1}{2}\sum_{n = -\infty}^{+\infty} \hat{f}(2\pi n)\, .\hspace{0.5in}\text{(1)}
$$
The Fourier transform of $f$ is:
\begin{align}
\hat{f}(q) &\;=\; \int_{-\infty}^{+\infty} dx\, e^{-i q x}\, \mathrm{erfc}(|x|)\\[0.1in]
&\;=\; 2\int_0^{+\infty} dx\, \cos(q x)\, \mathrm{erfc}(x)\hspace{0.5in}\text{Since $\mathrm{erfc}(|x|)$ is even}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dx\, \cos(q x)\,\int_x^{\infty}dt\, e^{-t^2} \hspace{0.5in}\text{Definition of $\mathrm{erfc}$}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\int_0^t dx\, \cos(q x) \hspace{0.5in}\text{Reverse order of integration}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\int_0^{+\infty} dt\, e^{-t^2}\,\frac{\sin(q t)}{q}\\[0.1in]
&\;=\; \frac{4}{\sqrt{\pi}}\frac{D(q/2)}{q}
\end{align}
In the last line above, "$D$" is the Dawson function according to the third definition here.
This expression for $\hat{f}(q)$ is valid everywhere except at $q = 0$, where
$\hat{f}(0) = 2/\sqrt{\pi}$. Plugging all of this into (1) results in:
$$
\sum_{n=0}^{\infty} \mathrm{erfc}(n)
\;=\; \frac{1}{2}\left[1 + \frac{2}{\sqrt{\pi}} \,+\,\frac{8}{\sqrt{\pi}}\sum_{n = 1}^{\infty} \frac{D(\pi n)}{2\pi n}\right]\, .
$$
Evaluating this numerically in Mathematica yields $1.16199904795$.
You can obtain a result for the second integral in terms of Owen's T-function which is defined as
$$
T(x, p) = \frac{1}{2\pi}\int_{0}^{p} \frac{e^{-\frac{1}{2} x^2 (1+t^2)}}{1+t^2} \, dt \quad \left(-\infty < x, p < +\infty\right).
$$
From this definition, a simple calculation using Feynman's trick tells us that
$\require{cancel}$
\begin{align*}
\frac{d}{d\beta} 2\sqrt{\frac{\pi}{\alpha}}T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)& = \cancel{2}\sqrt{\frac{\pi}{\alpha}} \frac{1}{\cancel{2}\pi}\int_{0}^{ \frac{1}{\sqrt{\alpha}}}\frac{\partial}{\partial \beta} \frac{e^{-\frac{1}{\cancel{2}}\frac{\cancel{2}\alpha}{\alpha+1} \beta^2 (1+t^2)}}{1+t^2} dt\\
& = \frac{1}{\sqrt{\pi \alpha}}\int_{0}^{ \frac{1}{\sqrt{\alpha}}} \frac{-2\frac{\alpha}{\alpha+1}\cancel{(1+t^2)}\beta e^{-\frac{\alpha}{\alpha+1}\beta^2 (1+t^2)}}{\cancel{1+t^2}} dt\\
& = -\frac{2\beta}{\sqrt{\pi }} \frac{\sqrt{\alpha}}{\alpha+1}e^{-\frac{\alpha}{\alpha+1}\beta^2}\int_{0}^{ \frac{1}{\sqrt{\alpha}}} e^{-\left(\sqrt{\frac{\alpha}{\alpha+1}}\beta t\right)^2} \, dt\\
& \overset{\color{blue}{u = \sqrt{\frac{\alpha}{\alpha+1}}\beta t}}{=} -\frac{\color{green}{2}\cancel{\beta}}{\color{green}{\sqrt{\pi }}} \frac{\cancel{\sqrt{\alpha}}}{\alpha+1}e^{-\frac{\alpha}{\alpha+1}\beta^2}\frac{\color{blue}{\sqrt{\alpha+1}}}{\color{blue}{\cancel{\beta}}\cancel{\color{blue}{\sqrt{\alpha}}}} \color{green}{\int_{0}^{\frac{\beta}{\sqrt{\alpha+1}}} e^{-u^2} \, du}\\
& = - \frac{1}{\sqrt{\alpha+1}}e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2}\color{green}{\text{erf}\left( \frac{\beta}{\sqrt{\alpha+1}}\right)} \tag{1}
\end{align*}
Now let's define a function $I(\beta)$ in terms of the integral we want
$$
I(\beta) := \int_{0}^{\infty} e^{-\alpha x^2} \text{erf}(x+\beta) \, dx
$$
by again using Feynman's trick we get
\begin{align*}
\frac{d}{d\beta}I(\beta) = \int_{0}^{\infty}e^{-\alpha x^2} \frac{\partial}{\partial \beta }\text{erf}(x+\beta) \, dx = \int_{0}^{\infty}e^{-\alpha x^2} \frac{2}{\sqrt{\pi}}e^{-(x+\beta)^2} \, dx = \frac{2}{\sqrt{\pi}}\int_{0}^{\infty}e^{-\left(\left(\color{blue}{\sqrt{\alpha+1}}\right)^2x^2 +2\color{purple}{\beta}x +\color{green}{\beta^2}\right)} \, dx
\end{align*}
where we used the fundamental theorem of calculus to differentiate the error function, along with its definition $\text{erf}(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x} e^{-t^2} \, dt$. Now, from equation $4$ in section $3.2$ of the tables you linked we know that
$$
\int_{0}^{\infty} e^{-(\color{blue}{a}^2 t^2 + 2\color{purple}{z} t + \color{green}{c})} \, dt = \frac{\sqrt{\pi}}{2a}\text{erfc}\left( \frac{z}{a}\right)e^{-\left(c-\frac{z^2}{a^2}\right)} = \frac{\sqrt{\pi}}{2a}e^{-\left(c-\frac{z^2}{a^2}\right)}-\frac{\sqrt{\pi}}{2a}\text{erf}\left( \frac{z}{a}\right)e^{-\left(c-\frac{z^2}{a^2}\right)}
$$
using that $\text{erfc}(x) = 1 - \text{erf}(x)$ by definition. We thus get that
$$
I'(\beta) = \frac{1}{\sqrt{\alpha+1}}e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2}- \frac{1}{\sqrt{\alpha+1}}\text{erf}\left( \frac{\beta}{\sqrt{\alpha+1}}\right)e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2} \tag{2}
$$
Using the definition of the error function, notice that
$$
\frac{d}{d\beta}\left[ \frac{1}{2} \sqrt{\frac{\pi}{\alpha}}\text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) \right]= \frac{1}{\sqrt{\alpha+1}}e^{-\left(1-\frac{1}{\alpha+1}\right)\beta^2}
$$
And thus, we can combine the previous equation with equations $(1)$ and $(2)$ to get that
$$
\frac{d}{d\beta}I(\beta) = \frac{d}{d\beta}\left[\frac{1}{2} \sqrt{\frac{\pi}{\alpha}}\text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) + 2\sqrt{\frac{\pi}{\alpha}}T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)\right]
$$
But since we have two functions whose derivatives are equal, the functions being differentiated must be equal up to some constant $\color{blue}{C}$:
$$
I(\beta) = \sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2} \text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) + 2T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)\right) + \color{blue}{C}
$$
Evaluating at $\beta =0$ we get
$$
I(0) = \sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2} \text{erf}\left(0\right) + 2T\left(0, \frac{1}{\sqrt{\alpha}} \right)\right) +C = \sqrt{\frac{\pi}{\alpha}}\left( \frac{\text{arctan}\left(\frac{1}{\sqrt{\alpha}} \right)}{\pi}\right) +C \tag{3}
$$
where we used the fact that $T(0, x)= \frac{1}{2\pi}\int_{0}^{x} \frac{1}{1+t^2} \, dt= \frac{\arctan(x)}{2\pi}$. If we now evaluate $I(0)$ using our original integral definition we can obtain the value for $C$ and complete the problem. We see that
$$
I(0) = \int_{0}^{\infty} e^{-\sqrt{\alpha}^2 x^2} \text{erf}(x) \, dx
$$
and using equation $2$ from section $4.3$ of the tables you linked we get
\begin{align}
I(0) = \frac{\sqrt{\pi}}{2\sqrt{\alpha}} - \frac{1}{\sqrt{\pi\alpha}}\arctan(\sqrt{\alpha}) = \cancel{\frac{\sqrt{\pi}}{2\sqrt{\alpha}}}-\frac{1}{\sqrt{\pi\alpha}}\left(\cancel{\frac{\pi}{2}} - \arctan\left(\frac{1}{\sqrt{\alpha}}\right) \right) =\frac{\arctan\left(\frac{1}{\sqrt{\alpha}}\right)}{\sqrt{\pi \alpha}} \tag{4}
\end{align}
where we used that $\arctan(x) + \arctan\left( \frac{1}{x}\right)= \frac{\pi}{2}$ since $ \alpha > 0$ in order for the integral to converge. Combining equations $(3)$ and $(4)$ we get that $C=0$, so we can finally conclude that for $a>0$ and $b \in \mathbb{R}$:
$$
\boxed{\int_{0}^{\infty} e^{-\alpha x^2} \text{erf}(x+\beta) \, dx = \sqrt{\frac{\pi}{\alpha}}\left(\frac{1}{2} \text{erf}\left(\sqrt{\frac{\alpha}{\alpha+1}} \beta\right) + 2T\left(\sqrt{\frac{2\alpha}{\alpha+1}}\beta, \frac{1}{\sqrt{\alpha}} \right)\right)}
$$
The integral you give in your question can be derived from the general case above. First, notice that
$$
\frac{1}{\pi}\int_{0}^{1} \frac{e^{-\frac{x^2}{2} (1+t^2)}}{1+t^2} \, dt=2T\left(x,1 \right) = \frac{1}{4} \text{erfc}\left( \frac{x}{\sqrt{2}}\right) \left(2- \text{erfc}\left( \frac{x}{\sqrt{2}}\right) \right)
$$
which can be proven by differentiating both sides of the equation by $x$ and using Feynman's trick. We thus get that
\begin{align}
\int_{0}^{\infty} e^{- x^2} \text{erf}(x+\beta) \, dx &= \sqrt{\frac{\pi}{1}}\left(\frac{1}{2} \text{erf}\left(\sqrt{\frac{1}{1+1}} \beta\right) + 2T\left(\sqrt{\frac{\cancel{2}}{\cancel{1+1}}}\beta, \frac{1}{\sqrt{1}} \right)\right)\\
&= \sqrt{\pi}\left(\frac{1}{2} \text{erf}\left( \frac{\beta}{\sqrt{2}}\right) + 2T\left(\beta, 1\right)\right)\\
&= \sqrt{\pi}\left(\frac{1}{2} \left(1-\text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)\right) + \frac{1}{4} \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right) \left(2- \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right) \right)\right)\\
&= \sqrt{\pi}\left(\frac{1}{2}-\cancel{\frac{1}{2}\text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)} + \cancel{\frac{1}{2} \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)}- \frac{1}{4} \text{erfc}\left( \frac{\beta}{\sqrt{2}}\right)\text{erfc}\left( \frac{\beta}{\sqrt{2}}\right) \right)\\
&= \frac{\sqrt{\pi}}{2}\left(1-\frac{1}{2}\text{erfc}^2\left(\frac{\beta}{\sqrt{2}}\right)\right)
\end{align}
As a final note, a nice complimentary identity I found whilst solving this problem is the integral
$$
\int_{-\infty}^{\infty} e^{-\alpha x^2} \text{erf}(\gamma x + \beta) \, dx = \sqrt{\frac{\pi}{\alpha}} \text{erf}\left( \frac{\beta \sqrt{\alpha}}{\sqrt{\gamma^2 +\alpha}}\right)
$$
which might also be useful to you.
Best Answer
The problem can be reduced to $$I= \frac{1}{\sqrt{\pi}} \left(\text{e}\pi - f(1,1) - \frac{1}{\text{e}} f(2,\infty) \right), $$ where $f(a,b) = \int_0^b \text{e}^{-ax^2} \frac{1}{1+x^2}\text{d}x$. These integrals look a bit more common, and following @Conreu's insight, they can actually be solved exactly (at least for the needed $b=1$ and $b=\infty$) -- see the second edit.
Now to show the above relation: Write $$I = \text{e} \int_{-1}^\infty \text{e}^{-x^2} \text{erf}(x+2)\text{d}x = \frac{2\text{e}}{\sqrt{\pi}} \int_{-1}^\infty \int_0^{x+2} \text{e}^{-x^2-y^2}\text{d}x\text{d}y.$$ Switching to polar coordinates with $x=r\cos\phi$ and $y=r\sin\phi$ gives $$I = \frac{2\text{e}}{\sqrt{\pi}} \int_{0}^\pi \int_0^{r(\phi)} \text{e}^{-r^2}r\text{d}r\text{d}\phi,$$ where the complexity of the integration region is hidden in $r(\phi)$. The inner integral is trivial, and we get $$I = \frac{\text{e}}{\sqrt{\pi}} \int_0^\pi \left(1-\text{e}^{-r(\phi)^2} \right)\text{d}\phi. $$ Looking at the boundaries, we have $$r(\phi) = \begin{cases} \infty & 0\le \phi < \pi/4 \\\\ \frac{2}{\sin(\phi)-\cos(\phi)} & \pi/4 < \phi < 3\pi/4 \\\\ \frac{1}{\cos(\phi)} & 3\pi/4 < \phi < \pi \end{cases}.$$ Thus, $$I = \frac{\text{e}}{\sqrt{\pi}} \left( \pi - \int_{3\pi/4}^\pi \text{e}^{-\sec(\phi)^2} \text{d}\phi - \int_{\pi/4}^{3\pi/4} \text{e}^{-4[\sin(\phi) - \cos(\phi)]^{-2}} \text{d}\phi \right).$$ Some more trigonometry and substituting $x=\tan(\phi)$ in the first integral and $x=\cot(\phi)$ in the second one then gives the form stated above. The numerical results for the three terms are \begin{align*} \text{e}\sqrt{\pi} &= 4.818029\dots,\\\\ \frac{1}{\sqrt{\pi}}f(1,1) &= \frac{1}{\sqrt{\pi}}\int_0^1 \text{e}^{-x^2} \frac{1}{1+x^2}\text{d}x = 0.349133\dots,\\\\ \frac{1}{\text{e}\sqrt{\pi}}f(2,\infty) &= \frac{1}{\text{e}\sqrt{\pi}}\int_0^\infty \text{e}^{-2x^2} \frac{1}{1+x^2}\text{d}x = 0.109611\dots, \end{align*} and $I=4.359285\dots$.
Edit: As requested, a few more details on the trigonometry: Note that $\sin(\phi)-\cos(\phi) = \sqrt{2} \sin(\phi-\pi/4)$, so the second integral is $$\int_{\pi/4}^{3\pi/4} \text{e}^{-2\sin(\phi-\pi/4)^{-2}} \text{d}\phi = \int_{0}^{\pi/2} \text{e}^{-2\csc(\phi)^{2}} \text{d}\phi.$$ Now use $\csc(\phi)^2=1+\cot(\phi)^2$ and substitute $x=\cot(\phi)$.
Edit2: Exact solutions for $f(a,b)$ for special $b$ (again, credit to @Conreu).
Let's start with $b=\infty$ and write $I(a)=f(a,\infty)$. Then $$I'(a) = -\int_0^\infty \text{e}^{-ax^2} \frac{x^2}{1+x^2}\text{d}x = I(a) - \int_0^\infty \text{e}^{-ax^2}\text{d}x = I(a) - \frac{1}{2} \sqrt{\frac{\pi}{a}}.$$ The solution of this ODE is of the form $I(a)=c(a)\text{e}^a$, and inserting this yields $c'(a)\text{e}^a = - \frac{1}{2} \sqrt{\frac{\pi}{a}}$, which can readily be integrated to give $$c(a)=c_0 - \frac{\pi}{2}\text{erf}(\sqrt{a}).$$ Thus, we find $I(a) = \left(c_0 - \frac{\pi}{2}\text{erf}(\sqrt{a}) \right) \text{e}^a$, and with $a=0$ giving $I(0)=\frac{\pi}{2}$, we finally have $$f(a,\infty) = \frac{\pi}{2} \text{erfc}(\sqrt{a}) \text{e}^a.$$ Similarly, for $I(a)=f(a,b=1)$, differentiating gives $$I'(a) = -\int_0^1 \text{e}^{-ax^2} \frac{x^2}{1+x^2}\text{d}x = I(a) - \int_0^1 \text{e}^{-ax^2}\text{d}x = I(a) - \frac{1}{2} \sqrt{\frac{\pi}{a}} \text{erf}(\sqrt{a}).$$ Using $\int \text{e}^{-x} \text{erf}(\sqrt{x}) \frac{1}{\sqrt{x}}\text{d}x = \frac{\sqrt{\pi}}{2} \text{erf}(\sqrt{x})$ and $I(0) = \frac{\pi}{4}$, we then find $$f(a,1) = \frac{\pi}{4} \left(1 - \text{erf}(\sqrt{a})^2 \right) \text{e}^a.$$ Combining all the results gives the exact answer \begin{align} I &= \frac{1}{\sqrt{\pi}} \left(\text{e}\pi - \frac{\pi}{4} \left(1 - \text{erf}(1)^2 \right) \text{e} - \frac{1}{\text{e}} \frac{\pi}{2} \text{erfc}(\sqrt{2}) \text{e}^2 \right) \\\\ &= \text{e}\sqrt{\pi}\left( \frac{3}{4} - \frac{1}{2} \text{erfc}(\sqrt{2}) + \frac{1}{4} \text{erf}(1)^2 \right) = 4.359285\dots \end{align} Edit3: Final edit, my intuition about the $f$-integrals was correct. These are actually known as Owen's T function, which are defined as $$T(h,b) = \frac{1}{2\pi} \int_0^b \text{e}^{-\frac{1}{2} h^2 (1+x^2)}\frac{1}{1+x^2} \text{d}x,$$ and we have $f(a,b) = 2\pi \text{e}^a T(\sqrt{2a}, b)$. The result then also follows from special cases of this function known in the literature.