The answers are $\boxed{f\equiv 0}$, $\boxed{f\equiv t}$, and $\boxed{f(x) = (t+1)x - t-2}.$ The first two clearly work, and the second one works because \begin{align*} (t+1)f(1+xy) - f(x+y) &= (t+1)[(t+1)(1+xy) - t-2] - [(t+1)(x+y) - t-2] \\ &= (t+1)^2(1+xy) - (t+1)(x+y) - t(t+2) \\ &= (t+1)^2xy - (t+1)(x+y) + 1 \\ &= ((t+1)x-1)((t+1)y-1) \\ &= ((t+1)(x+1)-(t+2))((t+1)(y+1)-(t+2)) \\ &= f(x+1)f(y+1). \end{align*}Now, we show that these are the only such functions that work. Assume that $f\not \equiv 0$; we will show that either $f\equiv t$ on $\mathbb{R}$ or $f(x) = (t+1)x-(t+2)$ for all $x\in \mathbb{R}$. Let $P(x, y)$ denote the assertion. Note that $$P(x, 0)\implies (t+1)f(1) - f(x) = f(x+1)f(1).$$Thus, if $s = f(1)$, then $$f(x) = st+s - sf(x+1). \text{ }[*]$$Setting $x = 1$ yields $$s = st + s - sf(2),$$so $sf(2)=st$, so either $s=0$ or $f(2)=t$. If $s = 0$, then by the above, $f\equiv 0$, which is a contradiction (we're assuming that $f\not \equiv 0$), so we must have $f(2)=t$. Now, note that since $f(1)=s$, we have from $[*]$ that $$f(0) = st + s - sf(1) = st - (s^2-s),$$$$f(-1) = st + s - sf(0) = st + s - s[st - (s^2-s)] = (s-s^2)t + (s^3-s^2+s),$$$$f(-2) = st + s - sf(-1) = st+s-s[(s-s^2)t+(s^3-s^2+s)] = (s^3-s^2+s)t - (s^4-s^3+s^2-s).$$Now, note that \begin{align*} P(-1, -1) &\implies (t+1)f(2) - f(-2) = f(0)^2 \\ &\implies t^2+t - [(s^3-s^2+s)t-(s^4-s^3+s^2-s)] = [st - (s^2-s)]^2 \\ &\implies t^2 - (s^3-s^2+s-1)t + (s^4-s^3+s^2-s) = s^2t^2-(2s^3-2s^2)t+(s^4-2s^3+s^2) \\ &\implies (1-s^2)t^2 + (s^3-s^2-s+1)t + (s^3-s) = 0 \\ &\implies (1-s^2)[t^2+(1-s)t - s] = 0 \\ &\implies (1-s^2)(t-s)(t+1) = 0. \\ \end{align*}Since $t\neq -1$, it follows that $(1-s^2)(t-s)=0$, so $s\in \{-1, 1, t\}$. We have three cases. \newline Case 1: $s = 1$. Then, $[*]$ implies that $$f(x) = t+1-f(x+1)$$for all $x\in \mathbb{R}$. This readily implies that $f(x) = f(x+2)$ for all $x\in \mathbb{R}$. Now, note that $$P(-1, y)\implies (t+1)f(1-y) - f(-1+y) = f(0)f(y+1).$$We have $f(0) = f(2) = t$, and we have that $f(-1+y) = f((-1+y)+2) = f(1+y)$. Thus, $$(t+1)f(1-y) - f(1+y) = tf(1+y),$$so since $t\neq -1$, $$f(1-y) = f(1+y).$$Now, note that $$P(-x, y)\implies (t+1)f(1-xy) - f(-x+y) = f(-x+1)f(y+1),$$but by the above, $f(1-xy) = f(1+xy)$ and $f(-x+1) = f(x+1)$, so $$(t+1)f(1+xy) - f(-x+y) = f(x+1)f(y+1).$$Subtracting this from $P(x, y)$ yields that $f(x+y) = f(-x+y)$ for all $x, y\in \mathbb{R}$, so from $y=x$, $f(2x)=f(0)=t$ for all $x\in \mathbb{R}$, i.e. $f\equiv t$. So, in this case, the only possible solution is $f\equiv t$ (if $t\neq 1$, then since $s = f(1)=1$, there are no solutions, however, if $t=1$, then $f\equiv t = 1$. But either way there are no other solutions beyond the ones claimed.) \newline Case 2: $s = t$. If $t = 1$, then refer to Case $1$. Otherwise, assume that $t\neq 1$. Note that $[*]$ implies that $$f(x) = t^2+t - tf(x+1).$$Thus, $$f(0) = t^2+t-tf(1) = t,$$as $f(1)=s=t$. Thus, $$P(-1, y)\implies (t+1)f(1-y) - f(-1+y) = tf(y+1).$$Now, note that by $[*]$, $$f(y) = t^2+t - tf(y+1),$$so by $[*]$ again, \begin{align*} f(-1+y) &= t^2+t - tf(y) \\ &= t^2+t - t[t^2+t-tf(y+1)] \\ &= t^2f(y+1) - t^3 + t. \end{align*}Therefore, we have that $$(t+1)f(1-y) = f(-1+y) + tf(y+1) = (t^2+t)f(y+1) - (t^3-t),$$so since $t\neq -1$, it follows that $$f(1-y) = tf(1+y) - (t^2-t).$$However, if we replace $y$ with $-y$, we have $$f(1+y) = tf(1-y) - (t^2-t).$$Therefore, we have that \begin{align*} f(1-y) &= tf(1+y) - (t^2-t) \\ &= t[tf(1-y) - (t^2-t)] - (t^2-t) \\ &= t^2f(1-y) - (t^3-t) \\ \end{align*}Therefore, $$(t^2-1)f(1-y) = t^3-t,$$so since $t\neq -1$, and $t\neq 1$ (discussed in the beginning of this case that we can redirect $t=1$ to Case $1$), we have that $f(1-y) = t$, for all $y\in \mathbb{R}$, so $f\equiv t$. \newline Case 3: $s = -1$. Then, $[*]$ implies that $$f(x) = f(x+1) - t-1,$$i.e. $$f(x+1) = f(x)+t+1,$$for all $x\in \mathbb{R}$. Therefore, $f(0) = f(1) - t - 1 = -t-2$. Now, note that $$P(-1, y)\implies (t+1)f(1-y) - f(-1+y) = (-t-2)f(y+1),$$and by $[*]$, we have $$f(-1+y) = f(y) - t-1 = [f(y+1)-t-1] - t-1 = f(y+1) - 2(t+1),$$so $$(t+1)f(1-y) - f(y+1) + 2(t+1) = (-t-2)f(y+1),$$so $$(t+1)f(1-y) + (t+1)f(1+y) = -2(t+1),$$so since $t\neq -1$, $$f(1-y) + f(1+y) = -2.$$Now, note that $$P(x, y)\implies (t+1)f(1+xy) - f(x+y) = f(x+1)f(y+1),$$and $$P(-x, y)\implies (t+1)f(1-xy) - f(-x+y) = f(-x+1)f(y+1).$$Summing these, we have that $f(1+xy) + f(1-xy) = f(x+1) + f(-x+1) = -2,$ so $$-2(t+1) - [f(x+y) + f(-x+y)] = -2f(y+1).$$Therefore, it follows that the quantity $$f(x+y) + f(-x+y)$$is independent of $x$. Therefore, if $r, s\in \mathbb{R}$, then we have that $$f\left(\frac{r-s}{2} + \frac{r+s}{2}\right) + f\left(-\frac{r-s}{2} +\frac{r+s}{2}\right) = f\left(\frac{r+s}{2} + \frac{r+s}{2}\right) + f\left(-\frac{r+s}{2} + \frac{r+s}{2}\right),$$i.e. $$f(r) + f(s) = f(r+s) + f(0).$$Thus, $$(f(r)-f(0)) + (f(s)-f(0)) = (f(r+s)-f(0)),$$so if $g(x) = f(x)-f(0) = f(x) +t+2,$ then $g$ is Cauchy. Now, note that $$P(-x, x)\implies (t+1)f(1-x^2) - f(0) = f(x+1)f(-x+1) = f(x+1)[-2-f(x+1)] \leq 1,$$so $$(t+1)f(1-x^2) \leq f(0)+1 = -t-1.$$Thus, if $t+1 > 0$, we have $$f(1-x^2) \leq -1,$$so $f$ is bounded above by $-1$ on the interval $[0, 1]$. If $t+1 < 0$, then $$f(1-x^2) \geq -1,$$so $f$ is bounded below by $-1$ on the interval $[0,1]$. (Since $t\neq -1$, $t+1=0$ is not a possible case). Thus, either way, $f$ is bounded in some direction on $[0, 1]$, so $g$ is bounded in some direction on $[0, 1]$. However, since $g$ is additive, this means that $g(x) = cx$ for some constant $c$, so $f(x) = g(x)-t-2 = cx - t-2$ for some constant $c$. Since $f(1) = s = -1$, it follows that $c - t - 2 = -1$, so $c = t + 1$. Thus, $$f(x) = (t+1)x - t-2.$$Thus, in all cases, we have that $f$ must be in our claimed solution set, as desired. $\blacksquare$