To show something is a group you need:
$1.$ Well defined binary operation on a nonempty set.
$2.$ The group operation to be associative.
$3.$ There to be an identity under the operation.
$4$. The set to be closed under the operation.
$5.$ Closure under inverses.
(This might be overkill but early on it is good to justify them all). Alright, here we go...
$1.$ The operation is binary (if you really want you can check well defined). $\mathbb{R}$ is clearly nonempty so our set should be nonempty.
$2$. Is the operation $*$ associative? You need $(a*b)*c=a*(b*c)$.
This operation is associative. Notice that $$ \begin{align} (a*b)*c&=(a+b+ab)*c\\&=(a+b+ab)+c+(a+b+ab)c\\&=a+b+ab+c+ac+ab+abc\\ &= a+(b+c+bc)+a(b+c+bc)\\&=a*(b+c+bc)\\&=a*(b*c)\end{align}$$
$3.$ Is there an identity? That is, is there an $e$ so that $a*e=a$ for all $a$?
Let $b=0$, then $a*0=a+0+a\cdot 0=a+0+0=a$ for any $a \in \mathbb{R}$.
$4.$ Check that $a*b \in \mathbb{R}$ and not $-1$
$a*b=a+b+ab$, but if $a,b \in \mathbb{R}$, then $a+b+ab \in \mathbb{R}$. Done! Now if $a,b \neq -1$, can $a*b=a+b+ab=-1$? Solving for $a$ would yield $a=-1$ and similarly for $b$, but this is impossible as $a,b \neq -1$, so $a*b=a+b+ab\neq -1$ so long as $a,b \neq -1$. Therefore, the set is closed under $*$.
$5.$ That is, if $a \in \mathbb{R}$, we need to find an element $i$ so that $a*i=e$, where $e$ is the identity of the group we found before.
The identity is $0$. We want $i$ such that $a*i=0$. Well, $a*i=a+i+ai=0$. Solving for $i$ gives, $i=-\frac{a}{a+1}$, so if $a\neq -1$, we can always find an inverse. Notice if we define $i$ as before then $$a*i=a+\frac{-a}{a+1}+a\frac{-a}{a+1}=\frac{(a^2+a)-a-a^2}{a+1}=0$$.
But then we've shown this set under $*$ is a group! Moreover, we can show this group is commutative:
$$
\begin{align}
a*b&=a+b+ab \\
&=b+a+ba \\
&=b*a
\end{align}
$$
because $a,b \in \mathbb{R}$ and addition and multiplication are commutative in $\mathbb{R}$.
Best Answer
Here $$\varphi^{-1}(K)=\{g\in G:\varphi(g)\in K\}. $$
The following is a solution to your exercise.
Since $\varphi(e_G)=e_K$, we have $e_G\in \varphi^{-1}(K)$, so $\varphi^{-1}(K)\neq\varnothing$. Also $\varphi^{-1}(K)\subseteq G$ by definition. Thus $\varphi^{-1}(K)$ is a nonempty subset of $G$.
Let $a, b\in\varphi^{-1}(K)$. Then there exist $x, y\in K$ such that $\varphi(a)=x$ and $\varphi(b)=y$. Consider $ab^{-1}$: we have $$\begin{align}\varphi(ab^{-1})&=\varphi(a)\varphi(b^{-1}) \\ &=x\varphi(b)^{-1}\\ &=xy^{-1},\end{align}$$ so $ab^{-1}=\varphi^{-1}(xy^{-1})\in \varphi^{-1}(K)$.
Hence $\varphi^{-1}(K)\le G$ by the one-step subgroup lemma.