Are there $f:\mathbb{R}^{+}\times\mathbb R^+\rightarrow\mathbb{R}^{+}$ associative and not conjugate to the addition?
Consider the function $$\color{#df0000}{\boxed{f : \Bbb R^+ \times \Bbb R^+ \to \Bbb R^+, \qquad f(x, y) := x + y + 1}} .$$
It is straightforward to verify that $f$ is associative and that the functions $x \mapsto f(x, a)$ and $y \mapsto f(a, y)$ are injective for all $a$ (they are increasing). Suppose $f$ is conjugate to addition via the (continuous, injective) function $\varphi$, that is, that $$\phantom{(\ast)} \qquad \boxed{\varphi(x) + \varphi(y) + 1 = \varphi(x + y)} . \qquad (\ast)$$ Since $\varphi$ is continuous and injective, it is strictly monotone, and $(\ast)$ forces $\varphi$ to be increasing. Thus, as $x \searrow 0$, $\varphi(x)$ decreases and is bounded below (by $0$), so $\lim_{x \searrow 0} \varphi(x)$ exists, and we can extend $\varphi$ uniquely to a continuous function $\tilde\varphi: \tilde X \to \tilde X$, $\tilde X := [0, \infty)$. We can likewise extend $f$ to a (continuous) function $\tilde f : \tilde X \times \tilde X \to \tilde X$ defined by the same rule as $f$, and by continuity $(\ast)$ still holds if we replace $\varphi$ with $\tilde \varphi$ therein. Taking $y = 0$ and canceling leaves $1 + \tilde \varphi(0) = 0$, which contradicts that $\tilde\varphi$ only takes on nonnegative values. So, $f$ is not conjugate to addition.
First of all, your definition of "conjugate" is highly unnatural because it requires $\varphi:\mathbb{R}\to\mathbb{R}$ instead of $\varphi:X\to X$. Note also that by your definition, multiplication on $\mathbb{R}^+$ is not conjugate to addition on $\mathbb{R}$, since you have only defined what it means for two operations on the same set to be conjugate. Moreover, your definition of conjugacy is not symmetric in $f$ and $g$, so it is unclear what it means. As stated, your definition essentially says that $f$ and $g$ are conjugate if the semigroup defined $g$ embeds in the semigroup defined by $f$. Taking $g$ to be addition, this would say that the semigroup defined by $f$ contains a copy of the addition semigroup $\mathbb{R}^+$, whereas it seems that what you intend is for $f$ to be contained in the addition semigroup $\mathbb{R}$.
Let me make a better definition which seems to be the meaning you are aiming for. A topological semigroup is a topological space $X$ together with a continuous associative binary operation. Two topological semigroups are isomorphic if there is a homeomorphism between them which turns the binary operation on one into the binary operation on the other. I will then say a binary operation on $\mathbb{R}^+$ is conjugate to addition if it is isomorphic as a topological semigroup to $(X,+)$ for some $X\subseteq\mathbb{R}$ (which is necesssarily an open interval).
Note in particular that there are three different isomorphism types of such topological semigroups $(X,+)$ where $X$ is an open interval: we could have $X=\mathbb{R}$, $X=\mathbb{R}^+$ (or equivalently $\mathbb{R}^-$), or $X=(a,\infty)$ for $a>0$ (or equivalently, $(-\infty,a)$ for $a<0$). These are not isomorphic since the first is a group, the second is not a group but $+$ is surjective, and in the third $+$ is not even surjective. And all of these are possible isomorphism types of topological semigroup structures on $\mathbb{R}^+$, since we can just pick a homeomorphism from $\mathbb{R}^+$ to each of these intervals and transport addition along the homeomorphism.
Now here's the interesting conclusion: every continuous associative binary operation on $\mathbb{R}^+$ which is injective in each variable is conjugate to addition in this sense. That is, every cancellative topological semigroup structure on $\mathbb{R}^+$ is isomorphic to one of $\mathbb{R}$, $\mathbb{R}^+$, or $(1,\infty)$ with addition.
Here's a proof. Since $\mathbb{R}^+$ is homeomorphic to $I=(0,1)$, I will work with $I$ instead. I will often write our binary operation $f$ as multiplication, so for instance $f(x,y)=xy$ (I will never use ordinary multiplication of elements of $I$).
Note first that since $f(x,\cdot)$ is injective, it is either increasing or decreasing. The set of $x$ such that $f(x,\cdot)$ is increasing is open by continuity, as is the set of $x$ such that $f(x,\cdot)$ is decreasing, so by connectedness $f(x,\cdot)$ is either increasing for all $x$ or decreasing for all $x$. But note that if we compose $f(x,\cdot)$ and $f(y,\cdot)$ we get $f(xy,\cdot)$ by associativity, and so $f(xy,\cdot)$ is always increasing. We conclude that actually $f(x,\cdot)$ is always increasing, and similarly $f(\cdot,x)$ is increasing.
Now fix some element $a\in I$. Let us first assume that $a^2>a$. Then since multiplication is increasing we get $a<a^2<a^3<\dots$. Let $R=\lim a^n$. If $R<1$, observe that $R^2=R$ by continuity, but also $aR=R$ by continuity, which is a contradiction since $a<R$. Thus $R=1$.
Now let's interpolate between the powers of $a$. For any $q=\frac{m}{n}$ where $m\geq n\geq 1$ are integers, note that by the intermediate value theorem there is $x\geq a$ such that $x^n=a^m$ (specifically, there is such an $x$ between $a^{\lfloor q\rfloor}$ and $a^{\lfloor q\rfloor+1}$). This $x$ is moreover unique since $x\mapsto x^n$ is strictly increasing. Define $a^q$ to be this unique $x$ (which, by uniqueness, does not depend on the denominator $n$ chosen for $q$). Note moreover that if $xy<yx$ then $x^2y^2<xyxy<yx^2y<yxyx<y^2x^2$, and similarly $x^ny^n<y^nx^n$ for all integers $n\geq 1$. Contrapositively, then, if $x^n$ and $y^n$ commute for some $n$, then $x$ and $y$ must also commute. We conclude that all the numbers of the form $a^q$ commute, and then it follows easily that $a\mapsto a^q$ is a strictly increasing semigroup homomorphism $\mathbb{Q}\cap[1,\infty)\to [a,1)$. For any $r\in[1,\infty)$, define $a^{r^-}=\sup\{a^q:q\leq r\}$ and $a^{r^+}=\inf\{a^q:q>r\}$. Say that $r$ is bad if $a^{r^-}\neq a^{r^+}$.
Note now that there can be only countably many bad $r$, since the open intervals $(a^{r^-},a^{r^+})$ are disjoint for distinct bad $r$. Moreover, if $s$ is good and $r$ is bad, then $r+s$ is also bad, since $a^{(r+s)^-}=a^{r^-}a^{s^-}$ and $a^{(r+s)^+}=a^{r^+}a^{s^+}$. But since there must be uncountably many good $s$, this means that if there is any bad $r$, there would be uncountably many bad $r$. Thus there are no bad $r$; that is, $a^{r^-}=a^{r^+}$ for all $r\in[1,\infty)$.
We can now define $a^r=a^{r^-}=a^{r^+}$ for each $r\in[1,\infty)$ and it is clear that $g(r)=a^r$ is a continuous strictly increasing map $g:[1,\infty)\to [a,1)$. By continuity, the image of $g$ is connected, and thus is all of $[a,1)$, so $g$ is a bijection. Since $g$ is a semigroup homomorphism on the rationals, it is a semigroup homomorphism everywhere by continuity.
We have thus shown that if $a^2>a$, then there is a unique increasing semigroup isomorphism $L_a:[a,1)\to[1,\infty)$ (the "base $a$ logarithm"). In particular, for any $b>a$, we also have $b^2>b$, and by uniqueness, $L_b(x)=L_a(x)/L_a(b)$ for all $x\in [b,1)$. Let $c=\inf\{a\in I:a^2>a\}$. Fixing some $a>c$, we can then extend $L_a$ to all of $(c,1)$ by defining $L_a(b)=L_a(x)/L_b(x)$ for any $x>\max(a,b)$. Letting $r=\inf\{1/L_a(b):b>c\}$, then this definition makes $L_a$ an increasing semigroup isomorphism $h_+:(c,1)\to(r,\infty)$. (Note that $\{a\in I:a^2>a\}$ may be empty in which case we just say $c=1$ and $r=\infty$ and all of this is vacuous.)
Similarly, letting $d=\sup\{a\in I:a^2<a\}$, we can define an increasing semigroup isomorphism $h_-:(0,d)\to (-\infty,s)$ for some $s<0$ (again this may be vacuous with $d=0$ and $s=-\infty$). If either $c=0$ or $d=1$ we are now done, since either $h_+$ or $h_-$ is defined on all of $I$. So, we assume $c>0$ and $d<1$. In that case, by continuity we can extend $h_+$ to $[c,1)\to [r,\infty)$ and $h_-$ to $(0,d]\to(-\infty,s]$ and they remain homomorphisms.
If $c=d$, we see we must have $r=s=0$ (since we must have $c^2\geq c$ and $d^2\leq d$ by continuity), and the element $c$ is an identity element for our multiplication. Now note that $f^{-1}(I\setminus\{c\})\subset I^2$ is disconnected, separated into $f^{-1}(0,c)$ and $f^{-1}(c,1)$. It follows that $f^{-1}(\{c\})$ must have more than one point and in fact must contain points of both $(c,1)\times(0,c)$ and $(0,c)\times(c,1)$. So there is $a>c$ and $b<c$ such that $ab=c$. For any $a'$ with $c<a'<a$, by continuity $a'$ must have a right inverse between $b$ and $c$. But we also have $a^nb^n=c$ for any $n$, and since every element of $(c,1)$ is less than some power of $a$, we conclude every $a>c$ has a right inverse. Similarly every $a>c$ has a left inverse, and every $b<c$ has a left and right inverse. Thus every element of $I$ has an inverse on each side, which implies every element of $I$ has a two-sided inverse.
That is, the elements less than $c$ are just the inverses of the elements greater than $c$. We can thus extend $h_+$ to an increasing isomorphism $I\to \mathbb{R}$ by defining $h_+(b)=-h_+(b^{-1})$ for $b<c$, and thus we have an isomorphism of topological semigroups from $I$ to $\mathbb{R}$ with addition.
The one remaining case is that $c\neq d$, in which case we must have $c>d$. We then have $e^2=e$ for all $e\in [d,c]$. In particular, $cd<c^2=c$ but $(cd)d=cd^2=cd$, which is a contradiction since right multiplication by $d$ is strictly increasing. So this last case is impossible.