This problem is a dual question of "Splitting a renewal process".
Assume we have a renewal process $P_1$ with inter-renewal distribution $p_1(x)$ and rate $\lambda_1 = \lim_{t \to \infty} \frac{N_1(t)}{t}$, where $N_1(t)$ is the total number of renewals of $P_1$ during $[0,t]$. We want to find another renewal process $P_2$ with rate $\lambda_2$ such that the super position of $P_1$ and $P_2$ is a renewal process.
If $P_1$ is Poisson, $P_2$ would be simply Poisson and the superposition of them is another Poisson and everything works. But here, $P_1$ is general, not only Poisson.
Also, if $P_1$ and $P_2$ are independent and $P_2$ is renewal, the superposition $P_1+P_2$ is renewal if and only if both $P_1$ and $P_2$ are Poisson, or have special distribution (discussed here).
The question remains for the general case. To solve that, I am thinking of considering building the superposition by as a renewal process with inter-renewal distribution $p_t(x)$ where $\frac{1}{\mathbb{E}[p_t(x)]}=\lambda_1+\lambda_2$. Now, we have $P_1(t)$ and $P_1(t)+P_2(t)$, I don't know how to find $P_2(t)$. Maybe we could take advantage of the fact that pdf of sum of two random variables is the convolution of their pdf. Any idea? or simpler method?
Along the lines of my last comment in your previous question, you could do a reverse convolution type thingy:
Suppose we have a renewal process $N(t)$ with rate $\lambda>0$. Let $X$ be a random variable with the same distribution as the inter-arrival times of $N(t)$. Then $E[X]=1/\lambda$. Fix $\epsilon>0$. We want to add another process $A(t)$ (not necessarily renewal, and not necessarily independent of $N(t)$) so that the sum process $N(t)+A(t)$ is a renewal process of rate $\lambda + \epsilon$.
Fix a probability $p \in (0,1]$. Suppose $X$ can be viewed as a geometric sum of positive iid random variables $Y$, so that $X = \sum_{i=1}^G Y_i$ where $G$ is independent and geometrically distributed with success probability $p$. Thus, $E[X]=E[Y]/p$. The process $N(t)$ can be viewed as having arrivals defined only at the end of the last $Y_i$ interval in every geometric sum. Let $A(t)$ be the arrival process with arrivals at the end of all the other $Y_i$ intervals. Then $N(t)+A(t)$ is a renewal process with rate $1/E[Y]$. We can choose $p$ to make the rate difference equal to $\epsilon$:
$$ \epsilon = \frac{1}{E[Y]} - \frac{1}{E[X]} = \frac{1}{pE[X]} - \frac{1}{E[X]} = \frac{1/p-1}{E[X]} $$ So choose $p = \frac{1}{\epsilon E[X] + 1}$.
Now it may not be possible to make such a decomposition of $X$. We want $X=\sum_{i=1}^G Y_i$. The moment generating function for $X$ gives: $$ E[s^X] = \sum_{i=1}^{\infty} (1-p)^{i-1}pE[s^X|G=i] = \sum_{i=1}^{\infty} (1-p)^{i-1}pE[s^Y]^i = \frac{pE[s^Y]}{1-(1-p)E[s^Y]} $$ where I am not worrying about region of convergence issues for simplicity. So this can be used to solve for the moment generating function $E[s^Y]$ in terms of the moment generating function $E[s^X]$.