Note that $\ln\left(\frac{\lambda+1}\lambda\right)=\ln(\lambda+1)-\ln\lambda$. Given this, we write
$$\int^{-1}_{-\infty} e^{\lambda t}(\ln(\lambda+1)-\ln\lambda)d\lambda=\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda-\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda$$
Let's first evaluate $\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda$ since it looks simpler.
We first substitute $u=\lambda t$, which means $\lambda=\frac ut$ and $d\lambda=\frac1t du$.
$$\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda=\frac1t\int^{-t}_{-\infty} e^u\ln\frac ut du=\frac1t\int^{-t}_{-\infty} e^u(\ln u-\ln t) du=\frac1t\left(\int^{-t}_{-\infty} e^u\ln u du-\int^{-t}_{-\infty}\ln te^udu\right)$$
$\int^{-t}_{-\infty}\ln te^udu$ is trivial. One can clearly see $\int^{-t}_{-\infty}\ln te^udu=\ln t\int^{-t}_{-\infty}e^udu=\ln t \left.e^u\right|^{-t}_{-\infty}=(\ln t\pm i\pi) e^{-t}$
$\int^{-t}_{-\infty} e^u\ln u du$ is trickier, but doable. We use integration by parts.
Let $v=\ln u$, $dw=e^u du$. Then $dv=\frac1u$, $w=e^u$.
Thus, we have
$$\int^{-t}_{-\infty} e^u\ln u du=\left. \ln u e^u\right|^{-t}_{-\infty}-\int^{-t}_{-\infty}\frac{e^u}u du$$
$\left.\ln u e^u\right|^{-t}_{-\infty}=\ln(-t)e^{-t}-\ln(-\infty)e^{-\infty}=(\ln t+i\pi)e^{-t}-\left(\ln\infty+i\pi\right) e^{-\infty}$. You can show that $\lim_{x\to\infty}\ln xe^{-x}=0$, so $(\ln t+i\pi)e^{-t}-\left(\ln\infty+i\pi\right) e^{-\infty}=(\ln t\pm i\pi)e^{-t}.$
$\int^{-t}_{-\infty}\frac{e^u}u du$ is the Exponential integral, which has no closed form. However, we can write $\int^{-t}_{-\infty}\frac{e^u}u du=\operatorname{Ei}(-t)$.
Combining the two integrals, we have
$$\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda=\frac1t\left((\ln t\pm i\pi)e^{-t}-\operatorname{Ei}(-t))-(\ln t\pm i\pi)e^{-t}\right)=-\frac1t\operatorname{Ei}(-t)$$
To evaluate $\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda$, we first write $u=\lambda+1$, which means $\lambda=u-1$ and $du=d\lambda$. Hence
$$\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda=\int^0_{-\infty} e^{(u-1)t}\ln udu=\int^0_{-\infty} e^{ut}e^{-t}\ln udu=e^{-t}\int^0_{-\infty} e^{ut}\ln udu$$
Does $\int^0_{-\infty} e^{ut}\ln udu$ look familiar to you? It should, since we evaluated this $\int^{-1}_{-\infty} e^{ut}\ln udu$ earlier. Using the same substitutions, but replacing $-1$ and $-t$ with $0$ in the upper bound, we arrive with
$$\int^0_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda=\frac1t\left(\ln u \left.e^u\right|^0_{-\infty}-\operatorname{Ei}(0)-\ln t \left.e^u\right|^0_{-\infty}\right)=\frac1t(\ln0-\operatorname{Ei}(0)-\ln t)$$
Inserting this back in,
$$e^{-t}\int^{-1}_{-\infty} e^{ut}\ln udu=\frac{e^{-t}}t(\ln0-\operatorname{Ei}(0)-\ln t)$$
$\ln0-\operatorname{Ei}(0)$ is undetermined as $\ln0=\operatorname{Ei}(0)=-\infty$, but we can squeeze some meaning out of it by interpreting it as $\lim_{x\to0}\ln x-\operatorname{Ei}(x)$ instead. To evaluate this, we first use the integral definitions of $\ln x$ $\operatorname{Ei}(x)$:
$$\lim_{x\to0}\ln x-\operatorname{Ei}(x)=\lim_{x\to0}\int^x_1\frac1zdz-\int^x_{-\infty}\frac{e^z}zdz$$
$$=\int^x_1\frac1zdz-\int^x_1\frac{e^z}zdz-\int^1_{-\infty}\frac{e^z}zdz$$
You may recall that $\int^b_a [f(x)-g(x)]dx=\int^b_a f(x)dx-\int^b_a g(x)dx$. The opposite also applies, i.e $\int^b_a f(x)dx-\int^b_a g(x)dx=\int^b_a [f(x)-g(x)]dx$. Knowing this, we can combine $\int^x_1\frac1zdz-\int^x_1\frac{e^z}zdz$:
$$\lim_{x\to0}\int^x_1\frac1zdz-\int^x_1\frac{e^z}zdz-\int^1_{-\infty}\frac{e^z}zdz=\lim_{x\to0}\int^x_1\frac{1-e^z}zdz-\operatorname{Ei}(1)$$
Now apply the sum definition of $e^z$:
$$\lim_{x\to0}\int^x_1\frac{1-e^z}zdz-\operatorname{Ei}(1)=\lim_{x\to0}\int^x_1\frac{1-\sum^\infty_{n=0}\frac{z^n}{n!}}zdz-\operatorname{Ei}(1)$$
$$=\lim_{x\to0}\int^x_1\frac{1-1-\sum^\infty_{n=1}\frac{z^n}{n!}}zdz-\operatorname{Ei}(1)$$
$$=\lim_{x\to0}\int^x_1\sum^\infty_{n=1}\frac{z^{n-1}}{n!}dz-\operatorname{Ei}(1)$$
$$=\sum^\infty_{n=1}\frac1{n!}\int^1_0z^{n-1}dz-\operatorname{Ei}(1)=\sum^\infty_{n=1}\frac1{n!}\cdot\frac1n-\operatorname{Ei}(1)$$
Interestingly, according to MathWorld $\operatorname{Ei}(x)=\gamma+\ln x+\sum^\infty_{n=1}\frac{x^n}{n\cdot n!}$. This means $\operatorname{Ei}(1)=\gamma+\ln 1+\sum^\infty_{n=1}\frac{1^n}{n\cdot n!}\Rightarrow \sum^\infty_{n=1}\frac1{n\cdot n!}=\operatorname{Ei}(1)-\gamma$. An interesting coincidence. Inserting this back:
$$\sum^\infty_{n=1}\frac1{n!}\cdot\frac1n-\operatorname{Ei}(1)=\operatorname{Ei}(1)-\gamma+\operatorname{Ei}(1)=-\gamma$$
Hence,
$$\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda=\frac{e^{-t}}t(-\gamma-\ln t)=-\frac{e^{-t}}t(\gamma+\ln t)$$
Thus,
$$\int^{-1}_{-\infty} e^{\lambda t}\ln(\lambda+1)d\lambda-\int^{-1}_{-\infty} e^{\lambda t}\ln\lambda d\lambda=-\frac{e^{-t}}t(\gamma+\ln t)+\frac{\operatorname{Ei}(-t)}t$$
(Note 1: If $\Gamma(x,y)$ refers to the upper incomplete gamma function, $\Gamma(0,t)=\int^\infty_t s^{-1}e^{-s} ds=\int^\infty_t\frac{e^{-s}}s ds$. Let $z=-s$, $-dz=ds$. $\int^\infty_t\frac{e^{-s}}s ds=\int^{-\infty}_{-t}\frac{e^z}{-z}-dz=-\int^{-t}_{\infty}\frac{e^z}zdz=-\operatorname{Ei}(-t)$. Hence, $\Gamma(0,t)=-\operatorname{Ei}(-t)$, and the $-\frac{\Gamma(0,t)}t$ that appears in Mathematica's answer is equivalent to the $\frac{\operatorname{Ei}(-t)}t$ that appears in my answer.)
(Note 2: Note that $\ln -t=\ln t+i\pi$ or $\ln -t=\ln t-i\pi$ depending on which branch of $\ln t$ you use, but considering $\frac{\lambda+1}{\lambda}$ is always nonnegative, $\ln\left(\frac{\lambda+1}{\lambda}\right)$ is always real and hence there should be no complex terms.)
