We say that
Definition A function $f:X\times X\rightarrow X$ is associative if $\forall a,b,c \in X\; f(f(a,b),c)=f(a,f(b,c))$
Definition Associative functions $f,g:X\times X\rightarrow X$ with $X\subseteq \mathbb{R}$ are said to be conjugate if there exists a continuous injective function $\varphi: \mathbb{R}\rightarrow
\mathbb{R}$ s.t. $f\circ (\varphi, \varphi)=\varphi\circ g $
For example the multiplication on the positive reals is conjugate to the sum as $\times \circ (\exp,\exp)=\exp \circ +$.
So a natural question would be:
Are there continuous $f:\mathbb{R}^{+}\times \mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ associative and not conjugate to the addition and such that $\forall x\in\mathbb{R}^{+}\; f(x,\cdot), f(\cdot,x)$ are injective?
Note that the last hypothesis was added to exclude trivial cases like the constant functions.
Best Answer
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.